Vectors & 3D Geometry PYQ — Page 3

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

The shortest distance between the lines $\dfrac{x-4}{1} = \dfrac{y-3}{2} = \dfrac{z-2}{-3}$ and $\dfrac{x+2}{2} = \dfrac{y-6}{4} = \dfrac{z-5}{-5}$ is :

The square of the distance of the point $P(5, 6, 7)$ from the line $\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4}$ is equal to:

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3\left|\left(3\vec{a} + 2\vec{b}\right)\right| + 4\left|\left(3\vec{a} - 2\vec{b}\right)\right|$ is :

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

Two adjacent sides of a parallelogram PQRS are given by $\vec{PQ} = \hat{j} + \hat{k}$ and $\vec{PS} = \hat{i} - \hat{j}$. If the side PS is rotated about the point P by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then $\sin^2\left(\dfrac{5\alpha}{2}\right) - \sin^2\left(\dfrac{\alpha}{2}\right)$ is equal to:

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

Let a vector $\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0$, make an obtuse angle with the vector $\overrightarrow{\mathrm{b}}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}$ and an angle $\theta, \frac{\pi}{6}<\theta<\frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$。

Let $\vec{a_k} = (\tan\theta_k)\hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot\theta_k)\hat{j}$, where $\theta_k = \dfrac{2^{k-1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}$, $n > 5$. Then the value of $\dfrac{\displaystyle\sum_{k=1}^{n}|\vec{a_k}|^2}{\displaystyle\sum_{k=1}^{n}|\vec{b_k}|^2}$ is _____.

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

Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3 \hat{i}+2 \hat{j}+2 \hat{k}$. If the distance of the point Q from the point $\mathrm{R}(1,3,3)$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :

Let A and B be two distinct points on the line $\mathrm{L}: \frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Both A and B are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point $(1,2,3)$ on the line L . If O is the origin, then $\overrightarrow{O A} \cdot \overrightarrow{O B}$ is equal to:

Let A, B, C be three points in $x y$-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\mathrm{a} \hat{i}+(1-\mathrm{a}) \hat{j}$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :

Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \quad \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let $G$ be the centroid of the triangle $A B C$. Then $6\left(|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right)$ is equal to ________

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

If the image of the point $\mathrm{P}(1,0,3)$ in the line joining the points $\mathrm{A}(4,7,1)$ and $\mathrm{B}(3,5,3)$ is $\mathrm{Q}(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

The perpendicular distance, of the line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}$ from the point $\mathrm{P}(2,-10,1)$, is :

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$, $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}$ be a vector such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4$. Then $|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{d}})|^2$ is equal to ______ .

Line $L_1$ passes through the point $(1,2,3)$ and is parallel to Z -axis. Line $\mathrm{L}_2$ passes through the point $(\lambda, 5,6)$ and is parallel to $y$-axis. Let for $\lambda=\lambda_1, \lambda_2, \lambda_2 \lt \lambda_1$, the shortest distance between the two lines be 3 . Then the square of the distance of the point $\left(\lambda_1, \lambda_2, 7\right)$ from the line $\mathrm{L}_1$ is

Let $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

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

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{C}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c}=5$, then $|\vec{c}|$ is equal to

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\overrightarrow{\mathrm{a}} \times \hat{\mathrm{d}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{d}}=1$, If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda \hat{d}+\mu \overrightarrow{\mathrm{c}}|^2$ is equal to _______ .