Vector Algebra-JEE-Mains-PYQ’s

Mathematics Questions

Let $\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$ . Let the projection of the vector $v = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\overrightarrow{AC}$ of the parallelogram $ABCD$ be of length one unit. If $\alpha, \beta$ , where $\alpha > \beta$ , be the roots of the equation $\lambda^2x^2 - 6\lambda x + 5 = 0$ , then $2\alpha - \beta$ is equal to

(A) 3

(B) 6

(D) 1

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For a triangle $ABC$ , let $p = \overrightarrow{BC}, q = \overrightarrow{CA}$ and $r = \overrightarrow{BA}$ . If $|p| = 2\sqrt{3}, |q| = 2$ and $\cos \theta = \frac{1}{\sqrt{3}}$ , where $\theta$ is the angle between $p$ and $q$ , then $|p \times (q - 3r)|^2 + 3|r|^2$ is equal to :

(A) 200

(B) 220

(D) 340

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Let $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}, \vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{b}$ . If $c \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ , then $|a + c|^2$ is equal to :

(A) 30

(B) 33

(D) 35

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Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$ . Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is :

(A) $\frac{18}{7}$

(B) $\frac{15}{7}$

(D) 3

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Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c} + \vec{d}| = \sqrt{29}$ and $\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$ . If $\lambda_1, \lambda_2$ ( $\lambda_1 > \lambda_2$ ) are the possible values of $(c + d) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$ , then the equation $K^2x^2 + (K^2 - 5K + \lambda_1)xy + \left(3K + \frac{\lambda_2}{2}\right)y^2 - 8x + 12y + \lambda_2 = 0$ represents a circle, for $K$ equal to :

(A) 4

(B) -1

(D) 1

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Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$ . Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$ . Then such a vector $\hat{c}$ is:

(A) $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$

(B) $\frac{1}{\sqrt{5}}(\hat{j} - 2\hat{k})$

(D) $\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k})$

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Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}| + |\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}| - |\vec{a}-\vec{b}|} = \sqrt{2} + 1$ . Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

(A) $2 + \sqrt{2}$

(B) $2 + 4\sqrt{2}$

(D) $1 + \sqrt{2}$

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Let the angle $\theta$ , $0 < \theta < \frac{\pi}{2}$ between two unit vectors $\vec{a}$ and $\vec{b}$ be $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$ . If the vector $\vec{c} = 3\vec{a} + 6\vec{b} + 9(\vec{a} \times \vec{b})$ , then the value of $9(\vec{c} \cdot \vec{a}) - 3(\vec{c} \cdot \vec{b})$ is

(A) 31

(B) 29

(D) 27

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Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$ , $\lambda > 0$ . The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$ . Let $\vec{v} = \vec{v}_1 + \vec{v}_2$ , where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$ . Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

(A) $\frac{23}{2}$

(B) $\frac{25}{2}$

(D) 14

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Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$ , $\vec{b} = -3\hat{i} + 2\hat{j} + 5\hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and $\vec{a} \cdot \vec{c} = 3$ . If $\vec{b} \times \vec{c} = \vec{d}$ , then $|\vec{a} \cdot \vec{d}|$ is equal to :

(A) 15

(B) 18

(D) 9

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Vector Algebra-JEE-Mains-PYQ’s

Mathematics Questions

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