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1lsg90h1z | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$$ be two vectors such that $$|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$$ and $$|\vec{b}|=4$$. If $$\vec{c}... | [{"identifier": "A", "content": "$$\\cos ^{-1}\\left(-\\frac{1}{\\sqrt{3}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\cos ^{-1}\\left(\\frac{2}{3}\\right)$$\n"}, {"identifier": "C", "content": "$$\\cos ^{-1}\\left(\\frac{2}{\\sqrt{3}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\cos ^{-1}\\left(-\\frac{\\... | ["D"] | null | <p>Given $$|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$$</p>
<p>$$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$$</p>
<p>Dot product with $$\overrightarrow{\mathrm{a}}$$ on both sides</p>
<p>$$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$$ ..... (1)</p>
<p>Dot prod... | mcq | jee-main-2024-online-30th-january-morning-shift |
luxwcrzd | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Between the following two statements:</p>
<p>Statement I : Let $$\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$$. Then the vector $$\vec{r}$$ satisfying $$\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$$ and $$\vec{a} \cdot \vec{r}=0$$ is of magnitude $$\sqrt{10}$$.</p>
<p>Statement II... | [{"identifier": "A", "content": "Both Statement I and Statement II are correct.\n"}, {"identifier": "B", "content": "Both Statement I and Statement II are incorrect.\n"}, {"identifier": "C", "content": "Statement I is correct but Statement II is incorrect.\n"}, {"identifier": "D", "content": "Statement I is incorrect b... | ["D"] | null | <p>$$\begin{aligned}
& \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\
& |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\
& \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\
& \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \v... | mcq | jee-main-2024-online-9th-april-evening-shift |
luxwe3dl | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$$, where $$\alpha$$ and $$\beta$$ are integers and $$\alpha \beta=-6$$. Let the values of the ordered pair $$(\alpha, \beta)$$, for which the area of the parallelogram of diagonals $$\vec{a}+\vec{b}$$ and $$\vec{b... | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "17"}] | ["C"] | null | <p>Area of parallelogram whose diagonals are $$\vec{a}+\vec{b}$$ and $$\vec{b}+\vec{c}$$ is</p>
<p>$$\begin{aligned}
& =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\
& =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\
& =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\bet... | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z4lq | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let three vectors ,$$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$$ form a triangle such that $$\vec{c}=\vec{a}-\vec{b}$$ and the area of the triangle is $$5 \sqrt{6}$$. If $$\alpha... | [{"identifier": "A", "content": "14"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "10"}] | ["A"] | null | <p>To solve this, let's start with the given vector equation:</p>
<p>$$\vec{c} = \vec{a} - \vec{b}$$</p>
<p>Given vectors are:</p>
<p>$$\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$$</p>
<p>$$\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$$</p>
<p>Then, the vector $$\overri... | mcq | jee-main-2024-online-9th-april-morning-shift |
luy6z5a5 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$$ and $$\overrightarrow{O C}=3 \vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\overrightarrow{O A}$$ and $$\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilat... | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "40"}] | ["C"] | null | <p>$$\begin{aligned}
& 6|\vec{a} \times \vec{b}|=15 \\
& \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3cizlj/32ca14d9-c7d9-4752-a76d-bc023959c231/e1a4ec70-1043-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cizlk.png?for... | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxdun | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\mathrm{ABC}$$ be a triangle of area $$15 \sqrt{2}$$ and the vectors $$\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}$$ and $$\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}>0$$.... | [] | null | 54 | <p>Area of triangle $$A B C=15 \sqrt{2}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\
& \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -7 \\
6 & d & -2
\end{array}\right| \\
& =(7 ... | integer | jee-main-2024-online-4th-april-morning-shift |
lv3vefae | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrig... | [{"identifier": "A", "content": "1600"}, {"identifier": "B", "content": "1618"}, {"identifier": "C", "content": "1627"}, {"identifier": "D", "content": "1609"}] | ["B"] | null | <p>$$\begin{aligned}
& \left.\begin{array}{l}
\vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\
\vec{b}=11 \hat{i}-\hat{j}+\hat{k}
\end{array}\right] \\
& \vec{a} \cdot \vec{b}=44+1+1=46 \\
& |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\
& (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\
& (2 \vec{a}+3 \vec{b... | mcq | jee-main-2024-online-8th-april-evening-shift |
lv5gt1qa | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$$ and $$\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$$ be three given vectors. If $$\vec{r}$$ is a vector such that $$\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$$ and $$\vec{r} \cdot(\vec{b}-\vec{c})=0$$, then $$\frac{|593 \ve... | [] | null | 569 | <p>$$\begin{aligned}
& \vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k} \\
& \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} \\
& \vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} \\
& \vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} \\
& (\vec{r}-(\vec{b}+\vec{c})) \times \vec{a}=0 \\
& \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \... | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v4g3o | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>If $$\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$$ and $$\mathrm{D}(-1,-4,-2)$$ are the vertices of a quadrilateral ABCD, then its area is :</p> | [{"identifier": "A", "content": "$$24 \\sqrt{7}$$\n"}, {"identifier": "B", "content": "$$48 \\sqrt{7}$$\n"}, {"identifier": "C", "content": "$$24 \\sqrt{29}$$\n"}, {"identifier": "D", "content": "$$12 \\sqrt{29}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgfpgex/f708b26f-9a53-4063-a888-8e61f5ee5884/8151b490-1776-11ef-bded-616e4abb66b9/file-1lwgfpgey.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgfpgex/f708b26f-9a53-4063-a888-8e61f5ee5884/8151b490-1776-11ef-bded-616e4abb66b9... | mcq | jee-main-2024-online-5th-april-morning-shift |
lv7v3oem | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overr... | [] | null | 30 | <p>$$(\vec{a}+2 \vec{b}) \times \vec{c}=3(\vec{c} \times \vec{a})$$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \vec{b} \times \vec{c}+2(\vec{a} \times \vec{c})=0 \\
& (\vec{b}+2 \vec{a}) \times \vec{c}=0 \\
& \vec{c}=\lambda(\vec{b}+2 \vec{a}) \\
& \vec{c} \cdot \vec{a}=130 \Rightarrow \lambda=1 \\
& \vec{c}=4 \hat{i... | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s2007 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $$\vec{a} \cdot \vec{c}=-29$$, then $$\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$$ is equal to:</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p>$$\begin{gathered}
(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\
\Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\
=\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\
\quad=\lambda(7 \hat{i}+8 \hat{j}) \\
\Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\
\quad \vec{... | mcq | jee-main-2024-online-5th-april-evening-shift |
lv9s20g2 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}|=2,|\vec{b}|=3$$ and $$\vec{a}=\vec{b} \times \vec{c}$$. If $$\alpha \in\left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c}-\vec{a}|^2$$ is equal to:</p> | [{"identifier": "A", "content": "124"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "121"}, {"identifier": "D", "content": "105"}] | ["A"] | null | <p>$$\begin{aligned}
& \vec{a}=\vec{b} \times \vec{c} \\
& |\vec{a}|=2,|\vec{b}|=3
\end{aligned}$$</p>
<p>$$\vec{a} \cdot \vec{b}=0$$ and $$\vec{a} \cdot \vec{c}=0$$</p>
<p>$$\begin{aligned}
& |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\
& =4+|\vec{c}|^2
\end{aligned}$$</p>
<p>$$\begin{aligned... | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294hw | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{5}$$\n"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\frac{2}{3}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\
& \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\
& \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -1 \\
1 & 1 & 0
\end{array}\right| \\
&=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\
&=\ha... | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb294j7 | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$$. If $$\overrightarrow{\mathrm{c}}$$ is a is vector such that $$|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrig... | [{"identifier": "A", "content": "$$\\frac{3}{2} \\sqrt{6}$$\n"}, {"identifier": "B", "content": "$$\\frac{9}{2}(6-\\sqrt{6})$$\n"}, {"identifier": "C", "content": "$$\\frac{9}{2}(6+\\sqrt{6})$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{2} \\sqrt{3}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\
& \left|\begin{array}{ccc}
i & j & k \\
6 & 1 & -1 \\
1 & 1 & 0
\end{array}\right|=i(1)-j(1)+k(5) \\
& =i-j+5 k \\
& |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\
& |\vec{c}-\vec{a}|=2 \sqrt{2} \\... | mcq | jee-main-2024-online-6th-april-evening-shift |
lvc57b8v | maths | vector-algebra | vector-or-cross-product-of-two-vectors-and-its-applications | <p>Let $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$$ and a vector $$\vec{c}$$ be such that $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$$. If $$\vec{a} \cdot \vec{c}=13$$, then $$(24-\vec{b} \cdot \vec{c})$$ is equal to _______.</p> | [] | null | 46 | <p>Let $$\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$$</p>
<p>Given $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$$</p>
<p>$$\begin{gathered}
\Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\
(\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b}
\end{gathe... | integer | jee-main-2024-online-6th-april-morning-shift |
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