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1l6dwfnc0 | maths | 3d-geometry | lines-and-plane | <p>Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the p... | [{"identifier": "A", "content": "$$\\frac{147}{2}$$"}, {"identifier": "B", "content": "96"}, {"identifier": "C", "content": "$$\\frac{32}{3}$$"}, {"identifier": "D", "content": "54"}] | ["C"] | null | Let $\langle a, b, c\rangle$ be direction ratios of plane containing
<br/><br/>
$$
\begin{aligned}
&\text { lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text { and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\
&\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text { (i) } \\\\
&\text { and } 3 a+7 b+8 c=0 \quad \ldots \text... | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6dxjgo4 | maths | 3d-geometry | lines-and-plane | <p>The line of shortest distance between the lines $$\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$$ and $$\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$$ makes an angle of $$\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$$ with the plane $$\mathrm{P}: \mathrm{a} x-y-z=0$$, $$(a>0)$$. If the image of the point $$(1,1,-5)$$ in the... | [] | null | 3 | DR's of line of shortest distance<br/><br/>
$$
\left|\begin{array}{lll}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
0 & 1 & 1 \\
2 & 2 & 1
\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}
$$<br/><br/>
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$<br/><br... | integer | jee-main-2022-online-25th-july-morning-shift |
1l6hzmtfd | maths | 3d-geometry | lines-and-plane | <p>The largest value of $$a$$, for which the perpendicular distance of the plane containing the lines $$
\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$$ and $$\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$$ from the point $$(2,1,4)$$ is $$\sqrt{3}$$, is _________.</p> | [] | null | 2 | <p>Normal to plane $$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & a & { - 1} \cr
{ - 1} & 1 & { - a} \cr
} } \right|$$</p>
<p>$$ = \widehat i(1 - {a^2}) - \widehat j( - a - 1) + \widehat k(1 + a)$$</p>
<p>$$ = (1 - a)\widehat i + \widehat j + \widehat k$$</p>
<p>$$\therefore$$... | integer | jee-main-2022-online-26th-july-evening-shift |
1l6hzyvwf | maths | 3d-geometry | lines-and-plane | <p>The plane passing through the line $$L: l x-y+3(1-l) z=1, x+2 y-z=2$$ and perpendicular to the plane $$3 x+2 y+z=6$$ is $$3 x-8 y+7 z=4$$. If $$\theta$$ is the acute angle between the line $$L$$ and the $$y$$-axis, then $$415 \cos ^{2} \theta$$ is equal to _____________.</p> | [] | null | 125 | <p>$$L:lx - y + 3(1 - l)z = 1$$, $$x + 2y - z = 2$$ and plane containing the line $$p:3x - 8y + 7z = 4$$</p>
<p>Let $$\overrightarrow n $$ be the vector parallel to L.</p>
<p>then $$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
l & { - 1} & {3(1 - l)} \cr
1 & 2 & { - ... | integer | jee-main-2022-online-26th-july-evening-shift |
1l6jdxtys | maths | 3d-geometry | lines-and-plane | <p>Let the line $$\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$$ intersect the plane containing the lines $$\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$$ and $$4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in \mathbb{R}$$ at the point $$P(\alpha, \beta, \gamma)$$. Then the value of $$\alpha+\beta+\gamma$$ equals _____________.</p> | [] | null | 12 | <p>Equation of plane containing the line $$4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3$$ can be written as</p>
<p>$$4ax - y + 5z - 7a + \lambda (2x - 5y - z - 3) = 0$$</p>
<p>$$(4a + 2\lambda )x - (1 + 5\lambda )y + (5 - \lambda )z - (7z + 3\lambda ) = 0$$</p>
<p>Which is coplanar with the line</p>
<p>$${{x - 4} \over 1} =... | integer | jee-main-2022-online-27th-july-morning-shift |
1l6kkntgj | maths | 3d-geometry | lines-and-plane | <p>If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt{2}}, \\pm \\,\\frac{1}{\\sqrt{2}}, 0$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{5}},-\\frac{2}{\\sqrt{5}}, 0$$"}, {"identifier": "D", "content": "$$\\frac{1}{2},-\\frac{\... | ["B"] | null | <p>$${P_1}:ax + by + 0z = 3$$, normal vector : $${\overrightarrow n _1} = (a,b,0)$$</p>
<p>$${P_2}:ax + by + cz = 0$$, normal vector : $${\overrightarrow n _2} = (a,b,c)$$</p>
<p>Vector parallel to the line of intersection $$ = {\overrightarrow n _1} \times {\overrightarrow n _2}$$</p>
<p>$${\overrightarrow n _1} \time... | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nnkfa5 | maths | 3d-geometry | lines-and-plane | <p>Let the lines <br/><br/>$$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$$ and <br/><br/>$$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$$ be coplanar <br/><br/>and $$\mathrm{P}$$ be the plane containing these two lines. <br/><br/>Then which of the following points does <b>NOT</b> lie on P?</p> | [{"identifier": "A", "content": "$$(0,-2,-2)$$"}, {"identifier": "B", "content": "$$(-5,0,-1)$$"}, {"identifier": "C", "content": "$$(3,-1,0)$$"}, {"identifier": "D", "content": "$$(0,4,5)$$"}] | ["D"] | null | <p>$${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$$,</p>
<p>through a point $${\overrightarrow a _1} \equiv (1,2,3)$$</p>
<p>parallel to $${\overrightarrow b _1} \equiv (\lambda ,1,2)$$</p>
<p>$${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$$</p>
<p>through a p... | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p20npe | maths | 3d-geometry | lines-and-plane | <p>If the foot of the perpendicular from the point $$\mathrm{A}(-1,4,3)$$ on the plane $$\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\sqrt{26}$$"}, {"identifier": "C", "content": "2$$\\sqrt{2}$$"}, {"identifier": "D", "content": "$$\\sqrt{14}$$"}] | ["B"] | null | <p>$$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$$ satisfies the plane $$P:2x + my + nz = 4$$</p>
<p>$$ - 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16$$ ...... (i)</p>
<p>Line joining $$A( - 1,\,4,\,3)$$ and $$\left( { - 2,{7 \over 2},{3 \over 2}} \right)$$ is perpendicular to $$P:2x + my + nz = 4... | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6p39yy3 | maths | 3d-geometry | lines-and-plane | <p>Let a line with direction ratios $$a,-4 a,-7$$ be perpendicular to the lines with direction ratios $$3,-1,2 b$$ and $$b, a,-2$$. If the point of intersection of the line $$\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$$ and the plane $$x-y+z=0$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamm... | [] | null | 10 | <p>Given $$a\,.\,3 + ( - 4a)( - 1) + ( - 7)2b = 0$$ ...... (1)</p>
<p>and $$ab - 4{a^2} + 14 = 0$$ ....... (2)</p>
<p>$$ \Rightarrow {a^2} = 4$$ and $${b^2} = 1$$</p>
<p>$$\therefore$$ $$L \equiv {{x + 1} \over 5} = {{y - 2} \over 3} = {z \over 1} = \lambda $$ (say)</p>
<p>$$\Rightarrow$$ General point on line is $$(5\... | integer | jee-main-2022-online-29th-july-morning-shift |
1ldo4up2x | maths | 3d-geometry | lines-and-plane | <p>Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\mathrm{d^{2}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{250}{83}$$"}, {"identifier": "B", "content": "$$\\frac{250}{82}$$"}, {"identifier": "C", "content": "$$\\frac{15}{53}$$"}, {"identifier": "D", "content": "$$\\frac{25}{83}$$"}] | ["A"] | null | Plane $P$, is passing through intersection of the two planes, so,
<br/><br/>$$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\\\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$$
<br/><br/>It is perpendicular with plane, $2 x+y-2+1=0$
<br/><br/>So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \la... | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldo7avkk | maths | 3d-geometry | lines-and-plane | <p>The point of intersection $$\mathrm{C}$$ of the plane $$8 x+y+2 z=0$$ and the line joining the points $$\mathrm{A}(-3,-6,1)$$ and $$\mathrm{B}(2,4,-3)$$ divides the line segment $$\mathrm{AB}$$ internally in the ratio $$\mathrm{k}: 1$$. If $$\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|$$... | [] | null | 10 | Plane: $8 x+y+2 z=0$
<br><br>Given line $\mathrm{AB}: \frac{\mathrm{x}-2}{5}=\frac{\mathrm{y}-4}{10}=\frac{\mathrm{z}+3}{-4}=\lambda$
<br><br>Any point on line $(5 \lambda+2,10 \lambda+4,-4 \lambda-3)$
<br><br>Point of intersection of line and plane
<br><br>$$
\begin{aligned}
& 8(5 \lambda+2)+10 \lambda+4-8 \lambda... | integer | jee-main-2023-online-1st-february-evening-shift |
1ldo7ffl8 | maths | 3d-geometry | lines-and-plane | <p>Let $$\alpha x+\beta y+\gamma z=1$$ be the equation of a plane passing through the point $$(3,-2,5)$$ and perpendicular to the line joining the points $$(1,2,3)$$ and $$(-2,3,5)$$. Then the value of $$\alpha \beta y$$ is equal to _____________.</p> | [] | null | 6 | Plane :
<br/><br/>$$
a(x-3)+b(y+2)+c(z-5)=0
$$
<br/><br/>Dr's of plane : $3 \hat{i}-\hat{j}-2 \hat{k}$
<br/><br/>$$
\begin{aligned}
& <3,-1,-2> \\\\
& P: 3(x-3)-1(y+2)-2(z-5)=0 \\\\
& 3 x-9-y-2-2 z+10=0 \\\\
& 3 x-y-2 z=1 \\\\
& \therefore \alpha=3, \beta=-1, \gamma=-2 \\\\
& \Rightarrow \alpha \beta \gamma=6
\end{alig... | integer | jee-main-2023-online-1st-february-evening-shift |
ldo7laxw | maths | 3d-geometry | lines-and-plane | Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to<br/><br/> the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the
intercept of $\mathrm{P}$<br/><br/> on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is : | [{"identifier": "A", "content": "$\\frac{6}{\\sqrt{14}}$"}, {"identifier": "B", "content": "$\\sqrt{14}$"}, {"identifier": "C", "content": "$\\sqrt{\\frac{2}{7}}$"}, {"identifier": "D", "content": "$\\sqrt{\\frac{7}{2}}$"}] | ["B"] | null | P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$
<br/><br/>L: $\frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$
<br/><br/>$\because \mathrm{P}$ is parallel to $\mathrm{L}$
<br/><br/>$\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$
<br/><br/>$\Rightarrow 3 \alpha_{1}+5\left(\... | mcq | jee-main-2023-online-31st-january-evening-shift |
ldo9cplr | maths | 3d-geometry | lines-and-plane | The foot of perpendicular from the origin $\mathrm{O}$ to a plane $\mathrm{P}$ which meets the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ is $(2, \mathrm{a}, 4), \mathrm{a} \in \mathrm{N}$. If the volume of the tetrahedron $\mathrm{OABC}$ is 144 unit$^{3}$, then which of the following points is... | [{"identifier": "A", "content": "$(3,0,4)$"}, {"identifier": "B", "content": "$(0,6,3)$"}, {"identifier": "C", "content": "$(0,4,4)$"}, {"identifier": "D", "content": "$(2,2,4)$"}] | ["A"] | null | Equation of Plane:
<br/><br/>$$
\begin{aligned}
& (2 \hat{i}+a \hat{j}+4 \hat{\mathrm{k}}) \cdot[(\mathrm{x}-2) \hat{\mathrm{i}}+(\mathrm{y}-\mathrm{a}) \hat{\mathrm{j}}+(\mathrm{z}-4) \hat{\mathrm{k}}]=0 \\\\
& \Rightarrow 2 \mathrm{x}+\mathrm{ay}+4 \mathrm{z}=20+\mathrm{a}^{2} \\\\
& \Rightarrow \mathrm{A} \equiv\l... | mcq | jee-main-2023-online-31st-january-evening-shift |
ldo9f5ge | maths | 3d-geometry | lines-and-plane | Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["A"] | null | Equation of Plane :
<br/><br/>$$
2(x-1)-3(y+1)-6(z+5)=0
$$
<br/><br/>or $2 x-3 y-6 z=35$
<br/><br/>$\Rightarrow$ Required distance $=$
$$
\frac{|2(3)-3(-2)-6(2)-35|}{\sqrt{4+9+36}}
$$ = 5 | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldptgy8w | maths | 3d-geometry | lines-and-plane | <p>Let the line $$L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$$ intersect the plane $$2 x+y+3 z=16$$ at the point
$$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1,-1,-3)$$ on the line $$L$$. If $$\alpha$$ is the area of triangle $$P Q R$$, then $$\alpha^{2}$$ is equal to __________.</p> | [] | null | 180 | $\quad L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=r$ (say)
<br/><br/>Let $P \equiv\left(2 r_{1}+1,-r_{1}, r_{1}+3\right)$
<br/><br/>$P$ lies on $2 x+y+3 z=16$
<br/><br/>$\therefore 2\left(2 r_{1}+1\right)+\left(-r_{1}-1\right)+3\left(r_{1}+3\right)=16$
<br/><br/>$r_{1}=1$
<br/><br/>$P \equiv(3,-2,4)$
<br/><br/... | integer | jee-main-2023-online-31st-january-morning-shift |
1ldptu28a | maths | 3d-geometry | lines-and-plane | <p>Let $$\theta$$ be the angle between the planes $$P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$$ and $$P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15$$. Let $$\mathrm{L}$$ be the line that meets $$P_{2}$$ at the point $$(4,-2,5)$$ and makes an angle $$\theta$$ with the normal of $$P_{2}$$. If $$\alpha$$ is t... | [] | null | 9 | $P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$
<br/><br/>$$
\begin{aligned}
& P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15 \\\\
& \text { then } \cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{1}{2} \\\\
& \begin{aligned}
& \therefore \theta=\frac{\pi}{3}, \text { Now } \alpha=\frac{\pi}{2}-\theta \\\... | integer | jee-main-2023-online-31st-january-morning-shift |
ldqv1su4 | maths | 3d-geometry | lines-and-plane | A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then : | [{"identifier": "A", "content": "$a+b+\\sqrt{2} c=1$"}, {"identifier": "B", "content": "$\\sqrt{2} a+b+c=1$"}, {"identifier": "C", "content": "$\\sqrt{2} a-b+c=1$"}, {"identifier": "D", "content": "$a+\\sqrt{2} b+c=1$"}] | ["D"] | null | <p>$$l = {1 \over 2},m = {1 \over {\sqrt 2 }},n = \cos \theta $$</p>
<p>$${l^2} + {m^2} + {n^2} = 1$$</p>
<p>$$ \Rightarrow {1 \over 4} + {1 \over 2} + {n^2} = 1 \Rightarrow {n^2} = {1 \over 4} \Rightarrow n = \, + \,{1 \over 2}$$</p>
<p>$$\theta$$ is acute $$\therefore$$ $$n = {1 \over 2}$$</p>
<p>$$\therefore$$ $$\ov... | mcq | jee-main-2023-online-30th-january-evening-shift |
ldqvn3td | maths | 3d-geometry | lines-and-plane | If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is : | [{"identifier": "A", "content": "$\\frac{17}{5}$"}, {"identifier": "B", "content": "$\\frac{6}{13}$"}, {"identifier": "C", "content": "$\\frac{13}{6}$"}, {"identifier": "D", "content": "$\\frac{5}{17}$"}] | ["C"] | null | <p>Let $$P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$$ & $$R(1,1,2)$$</p>
<p>$$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$$</p>
<p>& $$\overrightarrow Q R = - \widehat i + (1 - k)\widehat j + 3\widehat k$$</p>
<p>$$\therefore$$ Normal to plane will be</p>
<p>$$\left| {\matrix{
{\widehat i} & {\... | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr7gixq | maths | 3d-geometry | lines-and-plane | <p>The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$\\frac{19}{3}$$"}, {"identifier": "D", "content": "$$\\frac{13}{3}$$"}] | ["A"] | null | <p>Equation of $${l_1} = {{x - 2} \over 2} = {{y - 6} \over 1} = {{z - 2} \over { - 2}}$$</p>
<p>Shortest distance with $${{x + 1} \over 2} = {{y + 4} \over { - 3}} = {z \over 2}$$ is</p>
<p>S.d $$ = \left| {{{\matrix{
3 & {10} & 2 \cr
2 & 1 & { - 2} \cr
2 & { - 3} & 2 \cr
} } \over {\left| { - 4\wideh... | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldr7v822 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the point $$(1,1,2)$$ and perpendicular to the line $$x-3 y+ 2 z-1=0=4 x-y+z$$ is $$\mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$$, then $$140(\mathrm{C}-\mathrm{B}+\mathrm{A})$$ is equal to ___________.</p> | [] | null | 15 | <p>Line of intersection of the planes $$x - 3y + 2z - 1 = 0$$ and $$4x - y + z = 0$$ is normal $$(\overrightarrow n )$$ to the required plane.</p>
<p>$$\overrightarrow n = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 3} & 2 \cr
4 & { - 1} & 1 \cr
} } \right| = - \widehat i ... | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsesbom | maths | 3d-geometry | lines-and-plane | <p>The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{17}{3}$$"}, {"identifier": "B", "content": "$$\\frac{97}{3}$$"}, {"identifier": "C", "content": "$$\\frac{16}{3}$$"}, {"identifier": "D", "content": "$$\\frac{73}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053/file-1let6xpto.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053... | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsffwjf | maths | 3d-geometry | lines-and-plane | <p>If the lines $${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$$ and $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :</p> | [{"identifier": "A", "content": "28"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "16"}] | ["A"] | null | <p>$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda $$ (say)</p>
<p>& $${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu $$ (say)</p>
<p>$$\therefore$$ $$\lambda + 1 = 2\mu + a$$ ...... (i)</p>
<p>$$2\lambda + 2 = 3\mu - 2$$ ..... (ii)</p>
<p>$$\lambda - 3 = \mu + 3$$ .... (iii... | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldswkl3l | maths | 3d-geometry | lines-and-plane | <p>Let the equation of the plane P containing the line $$x+10=\frac{8-y}{2}=z$$ be $$ax+by+3z=2(a+b)$$ and the distance of the plane $$P$$ from the point (1, 27, 7) be $$c$$. Then $$a^2+b^2+c^2$$ is equal to __________.</p> | [] | null | 355 | The line $\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$
<br/><br/>
$\because$ the plane $a x+b y+3 z=2(a+b)$
<br/><br/>
$\Rightarrow \mathrm{b}=2 \mathrm{a}$
<br/><br/>
$\&$ dot product of d.r.'s is zero
<br/><br/>
$\therefore \mathrm{a}-2 \mathrm{~b}+3=0$ $\therefore \mathrm... | integer | jee-main-2023-online-29th-january-morning-shift |
1ldv2zcrz | maths | 3d-geometry | lines-and-plane | <p>Let the equation of the plane passing through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$ be $$ax + by + cz = 65$$. Then the distance of the point (a, b, c) from the plane $$2x + 2y - z + 16 = 0$$ is ____________.</p> | [] | null | 9 | Let the equation of the plane is
<br/><br/>
$(x-2 y-z-5)+\lambda(x+y+3 z-5)=0$
<br/><br/>
$\because \quad$ it's parallel to the line
<br/><br/>
$x+y+2 z-7=0=2 x+3 y+z-2$ <br/><br/>So, vector along the line $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|$
<br/><br/>
$$
=-... | integer | jee-main-2023-online-25th-january-morning-shift |
1ldww23tv | maths | 3d-geometry | lines-and-plane | <p>If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\alpha,\beta,\gamma$$), then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "5"}] | ["D"] | null | Direction of line
<br/><br/>
$$
\begin{aligned}
\vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
0 & 3 & -1
\end{array}\right| \\\\
& =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\\\
& =-5 \hat{i}+\hat{j}+3 \hat{k}
\end{aligned}
$$
<br/><br/>
Equation of line
<br/><br/>
$$
\frac{x-3}{-5}=\frac{y-... | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldwwhcmx | maths | 3d-geometry | lines-and-plane | <p>Let the plane containing the line of intersection of the planes <br/><br/>P<sub>1</sub> : $$x+(\lambda+4)y+z=1$$ and <br/><br/>P<sub>2</sub> : $$2x+y+z=2$$ <br/><br/>pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of <br/><br/>the point (2$$\lambda,\lambda,-\lambda$$) from the plane P<sub>2</sub> ... | [{"identifier": "A", "content": "$$2\\sqrt6$$"}, {"identifier": "B", "content": "$$3\\sqrt6$$"}, {"identifier": "C", "content": "$$4\\sqrt6$$"}, {"identifier": "D", "content": "$$5\\sqrt6$$"}] | ["B"] | null | Equation of plane passing through point of intersection of $\mathrm{P} 1$ and $\mathrm{P} 2$<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_1+\mathrm{kP}_2 = 0 \\\\
& (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0
\end{aligned}
$$<br/><br/>
Passing through $(0,1,0)$ and $... | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldybfknv | maths | 3d-geometry | lines-and-plane | <p>The distance of the point ($$-1,9,-16$$) from the plane <br/><br/>$$2x+3y-z=5$$ measured parallel to the line <br/><br/>$${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$ is :</p> | [{"identifier": "A", "content": "13$$\\sqrt2$$"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "20$$\\sqrt2$$"}, {"identifier": "D", "content": "31"}] | ["B"] | null | Given, $${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$$
<br><br>$$ \Rightarrow $$ $${{x + 4} \over 3} = {{y - 2} \over -4} = {{z - 3} \over {12}}$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leyorqpr/6043abe7-b1a8-4575-ada8-a5ec43b191a3/29ee5170-bd24-11ed-8df0-25e8b2ace386/... | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnyni4l | maths | 3d-geometry | lines-and-plane | Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be<br/><br/> parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point<br/><br/> $\mathrm{A}(8,-1,-19)$ from the plane $\mathrm{P}$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$<br/><b... | [] | null | 26 | Plane $\mathrm{P} \equiv \mathrm{P}_1+\lambda \mathrm{P}_2=0$<br><br>
$$
\begin{aligned}
& (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\\\
& (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\\\
& \overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{b}}=0 \text { where } \overrightarrow{\mathrm... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgowhqoz | maths | 3d-geometry | lines-and-plane | <p>The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :</p> | [{"identifier": "A", "content": "$$(0,5,-2)$$"}, {"identifier": "B", "content": "$$(2,0,1)$$"}, {"identifier": "C", "content": "$$(1,-2,1)$$"}, {"identifier": "D", "content": "$$(-2,5,0)$$"}] | ["D"] | null | <p>The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is ... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpyb2eu | maths | 3d-geometry | lines-and-plane | <p>The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$3 \\sqrt{5}$$"}, {"identifier": "C", "content": "$$2 \\sqrt{6}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{5}$$"}] | ["C"] | null | 1. Determine the line of shortest distance between the given two lines:
<br/><br/>Direction vector of line 1: $$\vec{d_1} = 2\hat{i} + \hat{k}$$
<br/><br/>Direction vector of line 2: $$\vec{d_2} = \hat{i} - \hat{j} + \hat{k}$$
<br/><br/>Now, let's find the cross product $$\vec{N} = \vec{d_1} \times \vec{d_2}$$
<br/... | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrgb0dt | maths | 3d-geometry | lines-and-plane | <p>Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\frac{\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\mathrm{P}$$, then $$35 \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "126"}, {"identifier": "B", "content": "105"}, {"identifier": "C", "content": "85"}, {"identifier": "D", "content": "90"}] | ["A"] | null | Equation of plane after rotation :
<br/><br/>$$
\begin{aligned}
& (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\\\
\Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\\\
& \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\\\
\Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\\\
\Rightarrow & 16+4 \l... | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgsw3mg4 | maths | 3d-geometry | lines-and-plane | <p>Let the line $$l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\alpha, \beta, \gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$$, then $$\alpha+2 \beta+6 \gamma$$ is equal to ________... | [] | null | 11 | $L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda} $
<br/><br/>$ P: x+2 y+3 z=4$
<br/><br/>Vector parallel to line : $\langle 1,2, \lambda\rangle=\bar{b}$
<br/><br/>Normal vector to plane $P:<1,2,3\rangle=\bar{n}$
<br/><br/>Angle between plane and line is $\theta$
<br/><br/>Then, $\sin \theta=\frac{<1,2, \lambda>\cdot... | integer | jee-main-2023-online-11th-april-evening-shift |
1lgvqjg9j | maths | 3d-geometry | lines-and-plane | <p>Let the foot of perpendicular from the point $$\mathrm{A}(4,3,1)$$ on the plane $$\mathrm{P}: x-y+2 z+3=0$$ be N. If B$$(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$$ is a point on plane P such that the area of the triangle ABN is $$3 \sqrt{2}$$, then $$\alpha^{2}+\beta^{2}+\alpha \beta$$ is equal to __________... | [] | null | 7 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8adf-57893cbbad41/file-6y3zli1lnkifvyt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8a... | integer | jee-main-2023-online-10th-april-evening-shift |
1lgxh5015 | maths | 3d-geometry | lines-and-plane | <p>Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\alpha,\beta,\gamma)$$, then $$\alpha\beta+\beta\gamma+\gamma\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "74"}, {"identifier": "C", "content": "76"}, {"identifier": "D", "content": "70"}] | ["B"] | null | Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and <br><br>$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$
[Given]
<br><br>Let, the other vertex is $C(x, y, z)$
<br><br>According to the question,
<br><br>$$
\begin{aligned}
& \frac{2+0+x}{3}=2 \\\\
&\Rightarrow x =4
\end{aligned}
$$
<br><br>$$
\begin{ali... | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgxsud7o | maths | 3d-geometry | lines-and-plane | <p>Let P be the point of intersection of the line $${{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :</p> | [{"identifier": "A", "content": "$${x^2} + 18x - 72 = 0$$"}, {"identifier": "B", "content": "$${x^2} - 18x - 72 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x + 72 = 0$$"}, {"identifier": "D", "content": "$${x^2} - 18x + 72 = 0$$"}] | ["D"] | null | Given, equation of line is
<br/><br/>$$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\\\
& \therefore x=3 k-3, y=k-2, z=1-2 k
\end{aligned}
$$
<br/><br/>Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$
<br/><br/>$$
\begin{aligned}
&\... | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgylgam9 | maths | 3d-geometry | lines-and-plane | <p>For $$\mathrm{a}, \mathrm{b} \in \mathbb{Z}$$ and $$|\mathrm{a}-\mathrm{b}| \leq 10$$, let the angle between the plane $$\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}$$ and the line $$l: x-1=\mathrm{a}-y=z+1$$ be $$\cos ^{-1}\left(\frac{1}{3}\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is ... | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "85"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "25"}] | ["C"] | null | We have, $\theta=\cos ^{-1} \frac{1}{3}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{1}{3} \\\\
& \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}
$$
<br/><br/>The given plane line and are
<br/><br/>$$
a x+y-z=b $$
<br/><br/>$$ x-1=a-y=z+1... | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgylmp2h | maths | 3d-geometry | lines-and-plane | <p>Let $$\mathrm{P}$$ be the plane passing through the line <br/><br/>$$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$$ and the point $$(2,4,-3)$$. <br/><br/>If the image of the point $$(-1,3,4)$$ in the plane P <br/><br/>is $$(\alpha, \beta, \gamma)$$ then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Equation of line : $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$
<br><br>Let $B \equiv(2,4,-3)$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyvtkfw/de189598-fad7-463c-a96b-e9cf1720e557/77a21fc0-5ba1-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvtkfx.png?format=png" data-orsrc="https://app-co... | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00mq7x | maths | 3d-geometry | lines-and-plane | <p>Let $$\lambda_{1}, \lambda_{2}$$ be the values of $$\lambda$$ for which the points $$\left(\frac{5}{2}, 1, \lambda\right)$$ and $$(-2,0,1)$$ are at equal distance from the plane $$2 x+3 y-6 z+7=0$$. If $$\lambda_{1} > \lambda_{2}$$, then the distance of the point $$\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lam... | [] | null | 9 | Since $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are equidistant
<br/><br/>from plane $2 x+3 y-6 z+7=0$
<br/><br/>$$
\begin{aligned}
& \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\\\
& \Rightarrow|5+... | integer | jee-main-2023-online-8th-april-morning-shift |
1lh216te7 | maths | 3d-geometry | lines-and-plane | <p>If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "14"}, {"identifier": "D", "content": "12"}] | ["C"] | null | Equation of plane intersection of two plane
<br/><br/>$$
\mathrm{P}_1+\lambda \mathrm{P}_2=0
$$
<br/><br/>$$
\mathrm{P}_1: 2 x-y+z=3, \text { and } \mathrm{P}_2: 4 x-3 y+5 z+9=0
$$
<br/><br/>Equation of any plane passing through the intersection of given planes is
<br/><br/>$(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$
<br/>... | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2ybxdl | maths | 3d-geometry | lines-and-plane | <p>Let the line $$\mathrm{L}$$ pass through the point $$(0,1,2)$$, intersect the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and be parallel to the plane $$2 x+y-3 z=4$$. Then the distance of the point $$\mathrm{P}(1,-9,2)$$ from the line $$\mathrm{L}$$ is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "$$\\sqrt{74}$$"}, {"identifier": "C", "content": "$$\\sqrt{69}$$"}, {"identifier": "D", "content": "$$\\sqrt{54}$$"}] | ["B"] | null | Given line, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{1}=\lambda$ .........(i)
<br><br>$\therefore$ Any point on the line $A(2 \lambda+1,3 \lambda+2, \lambda+3)$
<br><br>Since, line $L$ pass through the point $B(0,1,2)$ and intersects line (i)
<br><br>$\therefore$ Direction ratio of $A B$ are
<br><br>$$
\begin{array}{ll}... | mcq | jee-main-2023-online-6th-april-evening-shift |
1lsgafa13 | maths | 3d-geometry | lines-and-plane | <p>Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\overrightarrow{\mathrm{BD}}=\hat{i}+2 \hat{j}+3 \hat{k}$$, then the area of the parallelogram is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2} \\sqrt{410}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{2} \\sqrt{306}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\sqrt{586}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{2} \\sqrt{474}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \text { Area }=|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}| \\
& =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
5 & -1 & 7 \\
1 & 2 & 3
\end{array}\right| \\
& =\frac{1}{2}|-17 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}|=\fra... | mcq | jee-main-2024-online-30th-january-morning-shift |
lvc57bf8 | maths | 3d-geometry | lines-and-plane | <p>If $$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)$$ and $$D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is</p> | [{"identifier": "A", "content": "$$\\frac{4 \\sqrt{2}}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{2 \\sqrt{2}}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 \\sqrt{2}}{3}$$\n"}, {"identifier": "D", "content": "$$2 \\sqrt{2}$$"}] | ["A"] | null | <p>$$A(3,1,-1), B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1), D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$$ are vertices of a quadrilateral</p>
<p>$$\begin{aligned}
& \overrightarrow{A C}=(2 \hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}-\hat{k}) \\
& =-\hat{i}+\hat{j}+2 \hat{k} \\
& \overrigh... | mcq | jee-main-2024-online-6th-april-morning-shift |
lCoOxMgGdxQ2IcRT | maths | 3d-geometry | lines-in-space | The lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar if : | [{"identifier": "A", "content": "$$k=3$$ or $$-2$$"}, {"identifier": "B", "content": "$$k=0$$ or $$-1$$"}, {"identifier": "C", "content": "$$k=1$$ or $$-1$$"}, {"identifier": "D", "content": "$$k=0$$ or $$-3$$"}] | ["D"] | null | Coplanar if
<br/><br/>$$\left| {\matrix{
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{l_1}} & {{m_1}} & {{n_1}} \cr
{{l_2}} & {{m_2}} & {{n_2}} \cr
} } \right| = 0$$
<br><br>$$\therefore$$ $$\left| {\matrix{
1 & { - 1} & { - 1} \cr
1 & 1 & { ... | mcq | aieee-2003 |
QB1hUjgch8hPRWNt | maths | 3d-geometry | lines-in-space | A line with direction cosines proportional to $$2,1,2$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$ . The co-ordinates of each of the points of intersection are given by : | [{"identifier": "A", "content": "$$\\left( {2a,3a,3a} \\right),\\left( {2a,a,a} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,2a} \\right)$$ "}, {"identifier": "D", "content": "$$\\le... | ["B"] | null | Let a point on the line
<br><br>$$x = y + a = z$$ is $$\left( {\lambda ,\lambda - a,\lambda } \right)$$
<br><br>and a point on the line
<br><br>$$x + a = 2y = 2z$$ is $$\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$$
<br><br>then direction ratio of the line joining these points are
<br><br>$$\lambda -... | mcq | aieee-2004 |
4OZMVKi7SP8wxWVL | maths | 3d-geometry | lines-in-space | If the straight lines
<br/>$$x=1+s,y=-3$$$$ - \lambda s,$$ $$z = 1 + \lambda s$$ and $$x = {t \over 2},y = 1 + t,z = 2 - t,$$ with parameters $$s$$ and $$t$$ respectively, are co-planar, then $$\lambda $$ equals : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$-2$$ "}] | ["D"] | null | The given lines are
<br><br>$$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$$
<br><br>and $$2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$$
<br><br>The lines are coplanar, if
<br><br>$$\left| {\matrix{
{0 - \left( { - 1} \right)} & { - 1 - 3} ... | mcq | aieee-2004 |
u6SMbm7CaVhVyKjm | maths | 3d-geometry | lines-in-space | The angle between the lines $$2x=3y=-z$$ and $$6x=-y=-4z$$ is : | [{"identifier": "A", "content": "$${0^ \\circ }$$ "}, {"identifier": "B", "content": "$${90^ \\circ }$$"}, {"identifier": "C", "content": "$${45^ \\circ }$$"}, {"identifier": "D", "content": "$${30^ \\circ }$$"}] | ["B"] | null | The given lines are $$2x = 3y = - z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$$ $$\,\,\,\left[ {} \right.$$ Dividing by $$6$$ $$\left. {} \right]$$
<br><br>and $$6x = - y = - 4z$$
<br><br>or $$\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$$ $$\,\,\,\,\left... | mcq | aieee-2005 |
Z4uIG8y06yV4jPYd | maths | 3d-geometry | lines-in-space | The two lines $$x=ay+b, z=cy+d;$$ and $$x=a'y+b' ,$$ $$z=c'y+d'$$ are perpendicular to each other if : | [{"identifier": "A", "content": "$$aa'+cc'=-1$$"}, {"identifier": "B", "content": "$$aa'+cc'=1$$ "}, {"identifier": "C", "content": "$${a \\over {a'}} + {c \\over {c'}} = - 1$$ "}, {"identifier": "D", "content": "$${a \\over {a'}} + {c \\over {c'}} = 1$$"}] | ["A"] | null | Equation of lines
<br><br>$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
<br><br>$${{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}$$
<br><br>Line are perpendicular
<br><br>$$ \Rightarrow aa' + 1 + cc' = 0$$ | mcq | aieee-2006 |
2cg5qE8eqUZZH2mZ | maths | 3d-geometry | lines-in-space | The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then | [{"identifier": "A", "content": "$$a=2,$$ $$b=8$$"}, {"identifier": "B", "content": "$$a=4,$$ $$b=6$$ "}, {"identifier": "C", "content": "$$a=6,$$ $$b=4$$"}, {"identifier": "D", "content": "$$a=8,$$ $$b=2$$"}] | ["C"] | null | Equation of line through $$\left( {5,1,a} \right)$$ and
<br><br>$$\left( {3,b,1} \right)$$ is $${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $$
<br><br>$$\therefore$$ Any point on this line is a
<br><br>$$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \... | mcq | aieee-2008 |
ynbPZFBIbyg94BX0 | maths | 3d-geometry | lines-in-space | If the straight lines $$\,\,\,\,\,$$ $$\,\,\,\,\,$$ $${{x - 1} \over k} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ $$\,\,\,\,\,$$ and$$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 3} \over k} = {{z - 1} \over 2}$$ intersects at a point, then the integer $$k$$ is equal to | [{"identifier": "A", "content": "$$-5$$"}, {"identifier": "B", "content": "$$5$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$-2$$"}] | ["A"] | null | The two lines intersect if shortest distance between them is zero $$i.e.$$
<br><br>$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$
<br><br>$$ \Righ... | mcq | aieee-2008 |
wccmQiRxzMjvLNB5 | maths | 3d-geometry | lines-in-space | <b>Statement - 1 :</b> The point $$A(1,0,7)$$ is the mirror image of the point
<br/><br>$$B(1,6,3)$$ in the line : $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$
<br/><br><b>Statement - 2 :</b> The line $${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ bisects the line
<br/><br>segment joining $$A(1,0... | [{"identifier": "A", "content": "Statement -1 is true, Statement -2 is true; Statement -2 is <b>not</b> a correct explanation for Statement -1."}, {"identifier": "B", "content": "Statement -1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false , Statement -2 is true."}, {"identifi... | ["A"] | null | The directions ratios of the line segment joining points
<br><br>$$A\left( {1,0,7} \right)\,\,$$ and $$\,\,\,B\left( {1,6,3} \right)$$ are $$0,6, - 4.$$
<br><br>The direction ratios of the given line are $$1,2,3.$$
<br><br>Clearly $$1 \times 0 + 2 \times 6 + 3 \times \left( { - 4} \right) = 0$$
<br><br>So, the given l... | mcq | aieee-2011 |
7cGbYtvbBGU0bEmH | maths | 3d-geometry | lines-in-space | If the line $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over 4}$$ and $${{x - 3} \over 1} = {{y - k} \over 2} = {z \over 1}$$ intersect, then $$k$$ is equal to : | [{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${9 \\over 2}$$"}, {"identifier": "D", "content": "$$0$$ "}] | ["C"] | null | Given lines in vector form are
<br><br>$$\overrightarrow r = \left( {\widehat i - \overrightarrow j + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i + 3\overrightarrow j + 4\overrightarrow j } \right)$$
<br><br>and $$\overrightarrow r = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\... | mcq | aieee-2012 |
CXIqtyKExltEbk3B | maths | 3d-geometry | lines-in-space | If the lines $${{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}$$ and $${{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}$$ are coplanar, then $$k$$ can have : | [{"identifier": "A", "content": "any value "}, {"identifier": "B", "content": "exactly one value"}, {"identifier": "C", "content": "exactly two values "}, {"identifier": "D", "content": "exactly three values"}] | ["C"] | null | Given lines will be coplanar
<br><br>If $$\,\,\,\,\left| {\matrix{
{ - 1} & 1 & 1 \cr
1 & 1 & { - k} \cr
k & 2 & 1 \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow - 1\left( {1 + 2k} \right) - \left( {1 + {k^2}} \right) + 1\left( {2 - k} \right) = 0$$
<br><br>$$ \Rightarrow k = 0,... | mcq | jee-main-2013-offline |
ikcTgrIigYUF6ly3PV0QM | maths | 3d-geometry | lines-in-space | The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and
<br/>$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval : | [{"identifier": "A", "content": "[0, 1)"}, {"identifier": "B", "content": "[1, 2)"}, {"identifier": "C", "content": "(2, 3]"}, {"identifier": "D", "content": "(3, 4]"}] | ["C"] | null | Shortest distance between the lines
<br><br>$${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$$
<br><br>and $${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$$ is
<br><br> $$\left| {{{\left| {\matrix... | mcq | jee-main-2016-online-9th-april-morning-slot |
wufSe9GWqPvxj5A6ZtUib | maths | 3d-geometry | lines-in-space | If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$
<br/><br/>and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to : | [{"identifier": "A", "content": "$${7 \\over 2}$$ "}, {"identifier": "B", "content": "$${2 \\over 7}$$"}, {"identifier": "C", "content": "$$-$$ $${7 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$ $${4 \\over 7}$$"}] | ["A"] | null | Let $$\theta $$ be the angle between the two lines
<br><br>Here direction cosines of $${x \over 2}$$ = $${y \over 2}$$ = $${z \over 1}$$ are 2, 2, 1
<br><br>Also second line can be written as :
<br><br>$${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$$
<br><br>$$ \therefore $$ its direc... | mcq | jee-main-2018-online-16th-april-morning-slot |
0mlTVuKdlxftvs27Ar3rsa0w2w9jwxk75cq | maths | 3d-geometry | lines-in-space | If the length of the perpendicular from the point ($$\beta $$, 0, $$\beta $$) ($$\beta $$ $$ \ne $$ 0) to the line,
<br/>$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$$ is $$\sqrt {{3 \over 2}} $$, then
$$\beta $$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "-1"}] | ["D"] | null | $${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)$$<br><br>
any point on line A = (p, 1, – p – 1)<br><br>
Now, DR of AP $$ \equiv $$ < p – $$\beta $$, 1 – 0, – p – 1 – $$\beta $$ ><br><br>
Which is perpendicular to line so<br><br>
(p – $$\beta $$). 1 + 0.1 – 1(– p... | mcq | jee-main-2019-online-10th-april-morning-slot |
SGw6KlnNG3CtzIgUGr18hoxe66ijvwv9w1o | maths | 3d-geometry | lines-in-space | The vertices B and C of a $$\Delta $$ABC lie on the line,
<br/><br/>$${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$$ such that BC = 5 units. <br/><br/>Then the
area (in sq. units) of this triangle, given that the
point A(1, –1, 2), is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$$5\\sqrt {17} $$"}, {"identifier": "C", "content": "$$\\sqrt {34} $$"}, {"identifier": "D", "content": "$$2\\sqrt {34} $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265608/exam_images/ei4rpd9as7gipkdqbmus.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264422/exam_images/cu7zggkrbhyehugwdzx8.webp"><img src="https://res.c... | mcq | jee-main-2019-online-9th-april-evening-slot |
BUAbYW0qzI2qZL65Fu5S1 | maths | 3d-geometry | lines-in-space | The length of the perpendicular from the point
(2, –1, 4) on the straight line,
<br/><br/>$${{x + 3} \over {10}}$$= $${{y - 2} \over {-7}}$$ = $${{z} \over {1}}$$
is :
| [{"identifier": "A", "content": "less than 2"}, {"identifier": "B", "content": "greater than 4\n"}, {"identifier": "C", "content": "greater than 2 but less than 3"}, {"identifier": "D", "content": "greater than 3 but less than 4"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266241/exam_images/fhdx0dohqlxs327d7ok9.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267798/exam_images/i5tycucchql2hd8qs3c1.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-8th-april-morning-slot |
IGkJuXOyBfS6DOS0NnzhG | maths | 3d-geometry | lines-in-space | If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then : | [{"identifier": "A", "content": "ab' + bc' + 1 = 0"}, {"identifier": "B", "content": "cc' + a + a' = 0"}, {"identifier": "C", "content": "bb' + cc' + ... | ["D"] | null | Equation of 1<sup>st</sup> line is
<br><br>$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$
<br><br>Dr's of 1<sup>st</sup> line = ($$a$$, 1 , c)
<br><br>Equation of 2<sup>nd</sup> line is
<br><br>$${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$$
<br><br>Dr's of 2<sup>nd</sup> line = ($$a'$$, c' ,... | mcq | jee-main-2019-online-9th-january-evening-slot |
RhDgRy3oAgbUFBBEoo7k9k2k5fommlo | maths | 3d-geometry | lines-in-space | If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($$\alpha $$, 7, 1)
is
$$\left( {{5 \over 3},{7 \over 3},{{17} \over 3}} \right)$$, then $$\alpha $$ is equal to______. | [] | null | 4 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267568/exam_images/jagkrfpsaoaaeae9ashz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - 3D Geometry Question 235 English Explanation">
<br><br>Dire... | integer | jee-main-2020-online-7th-january-evening-slot |
H7fMwfkoz6SPD9jgNn7k9k2k5gznyfy | maths | 3d-geometry | lines-in-space | The shortest distance between the lines
<br/><br/>$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and
<br/><br/>$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${7 \\over 2}\\sqrt {30} $$"}, {"identifier": "C", "content": "$$3\\sqrt {30} $$"}, {"identifier": "D", "content": "$$2\\sqrt {30} $$"}] | ["C"] | null | $$\overrightarrow a $$
= < 3, 8, 3 >
<br><br>$$\overrightarrow b $$
= < – 3, – 7, 6 >
<br><br>$$\overrightarrow p $$
= < 3, – 1, 1 >
<br><br>$$\overrightarrow q $$
= < –3, 2, 4 >
<br><br>$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} &... | mcq | jee-main-2020-online-8th-january-morning-slot |
HngSHjDPG5hOAlG07kjgy2xukfg75pwu | maths | 3d-geometry | lines-in-space | If (a, b, c) is the image of the point (1, 2, -3) in<br/><br/> the line $${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$$, then a + b + c is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "-1"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267424/exam_images/kdcf95puxmjkf83bykfg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Mathematics - 3D Geometry Question 223 English Explanation">
<br><br>Eq... | mcq | jee-main-2020-online-5th-september-morning-slot |
cy8c71WuyhUNtGF5jA1klrkomrx | maths | 3d-geometry | lines-in-space | Let a, b$$ \in $$R. If the mirror image of the point P(a, 6, 9) with respect to the line <br/><br/>$${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then | a + b |, is equal to : | [{"identifier": "A", "content": "88"}, {"identifier": "B", "content": "90"}, {"identifier": "C", "content": "86"}, {"identifier": "D", "content": "84"}] | ["A"] | null | Given, P(a, 6, 9)<br/><br/>Equation of line $${{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}$$<br/><br/>Image of point P with respect to line is point Q(20, b, $$-$$a $$-$$9)<br/><br/>Mid-point of P and Q = $$\left( {{{a + 20} \over 2},{{6 + b} \over 2},{{ - a} \over 2}} \right)$$<br/><br/>This point li... | mcq | jee-main-2021-online-24th-february-evening-slot |
Jp12LbScNdtH0VokX81klrmjwqw | maths | 3d-geometry | lines-in-space | Let $$\lambda$$ be an integer. If the shortest distance between the lines <br/><br/>x $$-$$ $$\lambda$$ = 2y $$-$$ 1 = $$-$$2z and x = y + 2$$\lambda$$ = z $$-$$ $$\lambda$$ is $${{\sqrt 7 } \over {2\sqrt 2 }}$$, then the value of | $$\lambda$$ | is _________. | [] | null | 1 | $${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$$<br><br>$${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$$ ....... (1)<br><br>Point on line = $$\left( {\lambda ,{1 \over 2},0} \right)$$<br><br>$${x \over 1} = {{y + 2\lambda } \over 1} = {{z -... | integer | jee-main-2021-online-24th-february-evening-slot |
L0USDlCO6DJtPBCMFT1kls3suts | maths | 3d-geometry | lines-in-space | The equation of the line through the point (0, 1, 2) and perpendicular to the line <br/><br/>$${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}$$ is : | [{"identifier": "A", "content": "$${x \\over 3} = {{y - 1} \\over { - 4}} = {{z - 2} \\over 3}$$"}, {"identifier": "B", "content": "$${x \\over 3} = {{y - 1} \\over 4} = {{z - 2} \\over { - 3}}$$"}, {"identifier": "C", "content": "$${x \\over { - 3}} = {{y - 1} \\over 4} = {{z - 2} \\over 3}$$"}, {"identifier": "D", "c... | ["C"] | null | $${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}} = \lambda $$<br><br>Any point on this line $$(2\lambda + 1,3\lambda - 1, - 2\lambda + 1)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264765/exam_images/f91a8wrnbdvn5tmzebxr.webp" style="max-width: 100%;height: auto;display... | mcq | jee-main-2021-online-25th-february-morning-slot |
8zfZrqb7fcL0HgCiZX1klt9uh4s | maths | 3d-geometry | lines-in-space | A line 'l' passing through origin is perpendicular to the lines<br/><br/>$${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br/><br/>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br/><br/>If the co-ordinates of the point in the firs... | [] | null | 44 | $${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$$<br><br>$${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $$ D.R. of $${l_1} = 1,2,2$$<br><br>$${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$$<br><br>$${l_2}... | integer | jee-main-2021-online-25th-february-evening-slot |
YtsOhjwJ46iodfXK1u1kmhvwg6d | maths | 3d-geometry | lines-in-space | Let the position vectors of two points P and Q be 3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$ and $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lin... | [{"identifier": "A", "content": "$$\\sqrt {171} $$"}, {"identifier": "B", "content": "$$\\sqrt {227} $$"}, {"identifier": "C", "content": "$$\\sqrt {482} $$"}, {"identifier": "D", "content": "$$\\sqrt {5} $$"}] | ["A"] | null | $$\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k$$ & $$\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k$$<br><br>$${\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\overrightarrow v _{QS}} = ( - 2,1, - 2)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265215/exam... | mcq | jee-main-2021-online-16th-march-morning-shift |
ASltiYdrFmn8DXdoyu1kmiwbg93 | maths | 3d-geometry | lines-in-space | If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, <i>l</i> $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L<sub>1</sub> and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 6 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 3}} $$"}] | ["A"] | null | (3, 5, 7) lie on given line L<sub>1</sub><br><br>$${{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}$$<br><br>$${{7 - b} \over 4} = 1 \Rightarrow b = 3$$<br><br>$${{3 - a} \over l} = 1 \Rightarrow 3 - a = l$$<br><br>M(4, 3, 8)<br><br>N(3, 5, 7)<br><br>DR'S of MN = (1, $$-$$2, 1)<br><br>MN $$ \bot $$ line L<sub>1</sub... | mcq | jee-main-2021-online-16th-march-evening-shift |
1krrpbbk7 | maths | 3d-geometry | lines-in-space | The lines x = ay $$-$$ 1 = z $$-$$ 2 and x = 3y $$-$$ 2 = bz $$-$$ 2, (ab $$\ne$$ 0) are coplanar, if : | [{"identifier": "A", "content": "b = 1, a$$\\in$$R $$-$$ {0}"}, {"identifier": "B", "content": "a = 1, b$$\\in$$R $$-$$ {0}"}, {"identifier": "C", "content": "a = 2, b = 2"}, {"identifier": "D", "content": "a = 2, b = 3"}] | ["A"] | null | Lines are $$x = ay - 1 = z - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$$ .... (i)<br><br>and $$x = 3y - 2 = bz - 2$$<br><br>$$\therefore$$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii)<br><br>$$\t... | mcq | jee-main-2021-online-20th-july-evening-shift |
1krti6m7i | maths | 3d-geometry | lines-in-space | If the shortest distance between the straight lines $$3(x - 1) = 6(y - 2) = 2(z - 1)$$ and $$4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R$$ is $${1 \over {\sqrt {38} }}$$, then the integral value of $$\lambda$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$1"}] | ["A"] | null | $${L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k$$<br><br>$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k$$<br><br><img src="https://res.clo... | mcq | jee-main-2021-online-22th-july-evening-shift |
1kto6ba4g | maths | 3d-geometry | lines-in-space | The distance of line $$3y - 2z - 1 = 0 = 3x - z + 4$$ from the point (2, $$-$$1, 6) is : | [{"identifier": "A", "content": "$$\\sqrt {26} $$"}, {"identifier": "B", "content": "$$2\\sqrt 5 $$"}, {"identifier": "C", "content": "$$2\\sqrt 6 $$"}, {"identifier": "D", "content": "$$4\\sqrt 2 $$"}] | ["C"] | null | $$3y - 2z - 1 = 0 = 3x - z + 4$$<br><br>$$3y - 2z - 1 = 0$$<br><br>D.R's $$\Rightarrow$$ (0, 3, $$-$$2)<br><br>$$3x - z + 4$$ = 0<br><br>D.R's $$\Rightarrow$$ (3, $$-$$1, 0)<br><br>Let DR's of given line are a, b, c<br><br>Now, 3b $$-$$ 2c = 0 & 3a $$-$$ c = 0<br><br>$$\therefore$$ 6a = 3b = 2c<br><br>a : b : c = 3... | mcq | jee-main-2021-online-1st-september-evening-shift |
1l55iy2cg | maths | 3d-geometry | lines-in-space | <p>Let the image of the point P(1, 2, 3) in the line $$L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$$ be Q. Let R ($$\alpha$$, $$\beta$$, $$\gamma$$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($$\alpha$$ + $$\beta$$ + $$\gamma$$) is equal to _________... | [] | null | 125 | <p>The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.</p>
<p>$$\therefore$$ Let a point on line be $$\lambda$$</p>
<p>$$ \Rightarrow {{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3} = \lambda $$</p>
<p>$$ \Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\... | integer | jee-main-2022-online-28th-june-evening-shift |
1l56ra7cv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is :</p> | [{"identifier": "A", "content": "$${{18} \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{22} \\over {3\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$${{46} \\over {3\\sqrt 5 }}$$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["A"] | null | <p>$${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$</p>
<p>$${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$</p>
<p>Now, $$\overrightarrow p \times \overrightarrow q = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & 3 & { - 1} \cr
2 & 1 & 3 \... | mcq | jee-main-2022-online-27th-june-evening-shift |
1l58a01dt | maths | 3d-geometry | lines-in-space | <p>If the two lines $${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$$ and $${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$$ are perpendicular, then an angle between the lines l<sub>2</sub> and $${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$$ is :</p> | [{"identifier": "A", "content": "$${\\cos ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$${\\sec ^{ - 1}}\\left( {{{29} \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\cos ^{ - 1}}\\left( {{2 \\over {29}}} \\right)$$"}, {"identifier": "D", "content": "$${\\cos ^{ - 1}}\\l... | ["B"] | null | <p>$$\because$$ L<sub>1</sub> and L<sub>2</sub> are perpendicular, so</p>
<p>$$3 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0$$</p>
<p>$$ \Rightarrow \alpha = 3$$</p>
<p>Now angle between l<sub>2</sub> and l<sub>3</sub>,</p>
<p>$$\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over ... | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58gby99 | maths | 3d-geometry | lines-in-space | <p>Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i - \widehat j + \widehat k$$ be three given vectors. Let $$\overrightarrow v $$ be a vector in the plane of $$\overrightarrow a $$ and $$\overrighta... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["D"] | null | <p>Let $$\overrightarrow v = {\lambda _1}\overrightarrow a + {\lambda _2}\overrightarrow b $$, where $${\lambda _1},\,{\lambda _2} \in R$$.</p>
<p>$$ = ({\lambda _1} + 2{\lambda _2})\widehat i + ({\lambda _1} - 3{\lambda _2})\widehat j + (2{\lambda _1} + {\lambda _2})\widehat k$$</p>
<p>$$\because$$ Projection of $$\... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59limf1 | maths | 3d-geometry | lines-in-space | <p>Let l<sub>1</sub> be the line in xy-plane with x and y intercepts $${1 \over 8}$$ and $${1 \over {4\sqrt 2 }}$$ respectively, and l<sub>2</sub> be the line in zx-plane with x and z intercepts $$ - {1 \over 8}$$ and $$ - {1 \over {6\sqrt 3 }}$$ respectively. If d is the shortest distance between the line l<sub>1</sub... | [] | null | 51 | <p>$${{x - {1 \over 8}} \over {{1 \over 8}}} = {y \over { - {1 \over {4\sqrt 2 }}}} = {z \over 0}\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ______L<sub>1</sub></p>
<p>or $${{x - {1 \over 8}} \over 1} = {y \over { - \sqrt 2 }} = {z \over 0}$$ ..... (i)</p>
<p>Equation of L<sub>2</sub></p>
<p>$${{x + {1 \over 8}} \over { - 6\sqrt 3 }}... | integer | jee-main-2022-online-25th-june-evening-shift |
1l5baeet6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$$ and $${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is $${1 \over {\sqrt 3 }}$$, then the sum of all possible value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}] | ["A"] | null | <p>Let $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k$$</p>
<p>$${\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k$$</p>
<p>$$\therefore$$ $$\ove... | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c2ach2 | maths | 3d-geometry | lines-in-space | <p>Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$$ and $${x \over 2} = {{y - 7} \over 3} = {z \over 1}$$ at the points A and B. Then (AB)<sup>2</sup> is equal to ___________.</p> | [] | null | 84 | Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$ <br/><br/>So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
<br/><br/>
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
<br/><br/>
$\Rightarrow 5 \lambda-3 \mu=-16$
<br/><br/>
An... | integer | jee-main-2022-online-24th-june-morning-shift |
1l5c2jfp5 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines <br/><br/>$$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$$ <br/><br/>and $$\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$... | [] | null | 2 | $\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1\end{array}\right|=-a \hat{i}-\hat{j}+(a-1) \hat{k}$ <br/><br/>$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec... | integer | jee-main-2022-online-24th-june-morning-shift |
1l5w1j95q | maths | 3d-geometry | lines-in-space | <p>Consider a triangle ABC whose vertices are A(0, $$\alpha$$, $$\alpha$$), B($$\alpha$$, 0, $$\alpha$$) and C($$\alpha$$, $$\alpha$$, 0), $$\alpha$$ > 0. Let D be a point moving on the line x + z $$-$$ 3 = 0 = y and G be the centroid of $$\Delta$$ABC. If the minimum length of GD is $$\sqrt {{{57} \over 2}} $$, then... | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3/file-1l6dmv120.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3... | integer | jee-main-2022-online-30th-june-morning-shift |
1l6f2sskl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x+7}{-6}=\frac{y-6}{7}=z$$ and $$\frac{7-x}{2}=y-2=z-6$$ is :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{29}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{37}{29}}$$"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{29}}{2}$$"}] | ["A"] | null | <p>$${L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}$$</p>
<p>Any point on it $${\overrightarrow a _1}( - 7,6,0)$$ and L<sub>1</sub> is parallel to $${\overrightarrow b _1}( - 6,7,1)$$</p>
<p>$${L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}$$</p>
<p>Any point on it $${\overrightarr... | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6gio0a1 | maths | 3d-geometry | lines-in-space | <p>The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2 y+3 z-5$$ is :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{21}{2}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{9}{2}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{73}{2}}$$"}, {"identifier": "D", "content": "1"}] | ["A"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbsbos.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650... | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6gjpve9 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be two points on the line $$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$$ at a distance $$\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.</p> | [] | null | 153 | <p>$$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nc1t7n/ab3c067b-907d-4d61-90d4-40c5dd219f33/4878de40-2c50-11ed-9dc0-a1792fcc650d/file-1l7nc1t7o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nc1t7n/a... | integer | jee-main-2022-online-26th-july-morning-shift |
1l6kkgbw1 | maths | 3d-geometry | lines-in-space | <p>If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$... | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "14"}] | ["B"] | null | <p>$$\because$$ PR is perpendicular to given line, so</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa9n3e/85e3a21e-e14a-4db6-947b-e473de8b2f29/a9e898a0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa9n3f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qa9n3e/85e3a21e-e14a-4... | mcq | jee-main-2022-online-27th-july-evening-shift |
1ldomd3je | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines <br/><br/>$${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$$ and <br/><br/>$${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$$ is :</p> | [{"identifier": "A", "content": "$$7\\sqrt 3 $$"}, {"identifier": "B", "content": "$$5\\sqrt 3 $$"}, {"identifier": "C", "content": "$$4\\sqrt 3 $$"}, {"identifier": "D", "content": "$$6\\sqrt 3 $$"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\
& \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\
& \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\
& \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\
& \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\
&... | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldpsmm9a | maths | 3d-geometry | lines-in-space | <p>Let the shortest distance between the lines
<br/><br/>$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and
<br/><br/>$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,
<br/><br/>then which of the following is NOT possible?</p> | [{"identifier": "A", "content": "$$\\alpha+2 \\gamma=24$$"}, {"identifier": "B", "content": "$$2 \\alpha+\\gamma=7$$"}, {"identifier": "C", "content": "$$\\alpha-2 \\gamma=19$$"}, {"identifier": "D", "content": "$$2 \\alpha-\\gamma=9$$"}] | ["A"] | null | $\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$
<br/><br/>$$
\begin{aligned}
& \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\\\
& \vec{a}_{1}=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}_{,} \vec{a}_{2}=-\hat{i}+\hat{j}+4 \hat{k} \\\\
& \vec{a}_{1}-\vec{a}_{2}=6 \hat{i}+(\lambda-1) \hat{j}-(\lambd... | mcq | jee-main-2023-online-31st-january-morning-shift |
ldr0kpgp | maths | 3d-geometry | lines-in-space | Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to : | [] | null | 158 | <p>$$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda $$</p>
<p>Any point on L can be taken as</p>
<p>$$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$$</p>
<p>Let $$A(5,3,8)$$</p>
<p>So, $$AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$$</p>
<p>$$[(\lambda - 3)\widehat i - \lambda \... | integer | jee-main-2023-online-30th-january-evening-shift |
1ldsfqrb6 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$$ and $${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$$ is :</p> | [{"identifier": "A", "content": "$$2\\sqrt3$$"}, {"identifier": "B", "content": "$$3\\sqrt3$$"}, {"identifier": "C", "content": "$$4\\sqrt3$$"}, {"identifier": "D", "content": "$$5\\sqrt3$$"}] | ["C"] | null | <p>$${\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k$$</p>
<p>$${\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k$$</p>
<p>$$\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k$$</p>
<p>$$\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k$$</p>
<p>S.D. $$ = {{\left| {\mat... | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsx6ir2 | maths | 3d-geometry | lines-in-space | <p>Let the co-ordinates of one vertex of $$\Delta ABC$$ be $$A(0,2,\alpha)$$ and the other two vertices lie on the line $${{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}$$. For $$\alpha \in \mathbb{Z}$$, if the area of $$\Delta ABC$$ is 21 sq. units and the line segment $$BC$$ has length $$2\sqrt{21}$$ ... | [] | null | 9 | <p>A. $\left(\mathrm{O}_{1} 2, \alpha\right)$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekt8kqs/c8535748-1d32-4d4a-8b11-6aa2b4c12893/a97e5c40-b582-11ed-b66a-d501f51666d6/file-1lekt8kqt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekt8kqs/c8535748-1d32-4d4a-8b11... | integer | jee-main-2023-online-29th-january-morning-shift |
1ldu4b5nl | maths | 3d-geometry | lines-in-space | <p>The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$$ is ($$\alpha,\beta,\gamma$$). Then, which of the following is NOT correct?</p> | [{"identifier": "A", "content": "$$\\frac{\\alpha}{\\beta}=-8$$"}, {"identifier": "B", "content": "$$\\frac{\\alpha \\beta}{\\gamma}=\\frac{4}{15}$$"}, {"identifier": "C", "content": "$$\\frac{\\beta}{\\gamma}=-5$$"}, {"identifier": "D", "content": "$$\\frac{\\gamma}{\\alpha}=\\frac{5}{8}$$"}] | ["C"] | null | <p>$$
\mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) }
$$</p><p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef5vgns/9610cbeb-a0ae-4272-89c8-8cbf8a31b61b/5f9aceb0-b267-11ed-9d4d-b96eca78f2e5/file-1lef5vgnx.png?format=png" data-orsrc="https://app-content.cdn.ex... | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu52orl | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$x+1=2y=-12z$$ and $$x=y+2=6z-6$$ is :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{5}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $L_{1}: \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$
<br/><br/>
$$
\begin{aligned}
& L_{2}: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\\\
& \text { S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\\\
& =\left|\frac{-2-6-6}{7}\right|=2 \text { unit... | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldu68afw | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$$ is $$\alpha$$, then 28$$\alpha^2$$ is equal to ____________.</p> | [] | null | 18 | Points $(1,2,3)$ and $(2,3,4)$
<br/><br/>
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
<br/><br/>
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
<br/><br/>
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
<br/><br/>
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$<br/><br/>
$$
\begin{aligned}
& \overrightarrow{a_{1}}-\... | integer | jee-main-2023-online-25th-january-evening-shift |
1ldv1rqg2 | maths | 3d-geometry | lines-in-space | <p>The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\sqrt{14}$$"}, {"identifier": "C", "content": "$$\\sqrt6$$"}, {"identifier": "D", "content": "$$2\\sqrt3$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/file-1lebz24vi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/fi... | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldv1wbyd | maths | 3d-geometry | lines-in-space | <p>Consider the lines $$L_1$$ and $$L_2$$ given by</p>
<p>$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$</p>
<p>$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.</p>
<p>A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ an... | [{"identifier": "A", "content": "$$4\\sqrt3$$"}, {"identifier": "B", "content": "$$2\\sqrt6$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$3\\sqrt2$$"}] | ["B"] | null | Let,
<br/><br/>
$$
\begin{aligned}
& P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2 \text {, } 3 \mu+3) \\\\
& \text { d.r's of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\\\
& \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-... | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwxg3ot | maths | 3d-geometry | lines-in-space | <p>If the shortest between the lines $${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$$ and $${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$$ is 6, then the square of sum of all possible values of $$\lambda$$ is :</p> | [] | null | 384 | Shortest distance between the lines<br/><br/> $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$<br/><br/> $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6<br/><br/>
Vector along line of shortest distance<br/><br/> $=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2... | integer | jee-main-2023-online-24th-january-evening-shift |
1ldybxlrv | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$ and $${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$$ is equal to ________</p> | [] | null | 14 | <p>For $${L_1}:$$$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$$</p>
<p>$$\therefore$$ Point on line is $$A(2,-1,6)$$</p>
<p>Parallel vector to this line is,</p>
<p>$$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$$</p>
<p>For $${L_2}:$$$${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \ov... | integer | jee-main-2023-online-24th-january-morning-shift |
lgnyep2q | maths | 3d-geometry | lines-in-space | Let $\mathrm{S}$ be the set of all values of $\lambda$, for which the shortest distance between<br/><br/> the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is 13. Then $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ is equal to : | [{"identifier": "A", "content": "306"}, {"identifier": "B", "content": "304"}, {"identifier": "C", "content": "308"}, {"identifier": "D", "content": "302"}] | ["A"] | null | Given the two lines :
<br/><br/>$$
\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} \\\\
\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}
$$
<br/><br/>Let's find the direction vectors of these lines: $\vec{d_1} = \langle 0, 4, 1 \rangle$ and $\vec{d_2} = \langle 3, -4, 0 \rangle$.
<br/><br/>Now, let's find the cross pro... | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgowajui | maths | 3d-geometry | lines-in-space | <p>The line, that is coplanar to the line $$\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{4}$$"}, {"identifier": "B", "content": "$$\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "C", "content": "$$\\frac{x-1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}$$"}, {"identifier": "D", "content": "$$\\frac{x+1}{1}=\\frac{y-2}{2}=... | ["B"] | null | <p>Given two lines:</p>
<p>$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$</p>
<p>and </p>
<p>$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$</p>
<p>These lines are coplanar if the determinant of the matrix</p>
<p>$$
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a... | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgrebnuy | maths | 3d-geometry | lines-in-space | <p>Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "8"}] | ["C"] | null | $$
\begin{aligned}
& (3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2)+\mu(\mathrm{x}-3 \mathrm{y}+2 \mathrm{z}-13)=0 \\\\
& 3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0 \\\\
& 9-4 \mu=0 \\\\
& \mu=\frac{9}{4} \\\\
& 4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0 \\\\
& -100+4 \alpha-54+18 \alpha=0 \\\\
& \Rightarrow \alpha=7 \\\\
& \text ... | mcq | jee-main-2023-online-12th-april-morning-shift |
1lguwwp0s | maths | 3d-geometry | lines-in-space | <p>Let a line $$l$$ pass through the origin and be perpendicular to the lines</p>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k... | [] | null | 5 | We have,<br/>
<p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p>
<p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.</p>
Let direction ratio of line $l$ be $a... | integer | jee-main-2023-online-11th-april-morning-shift |
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