Vectors & 3D Geometry PYQ — Page 2

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ 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:

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 :

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 _____.

Let $P$ be a point in the plane of the vectors $\overrightarrow{A B}=3 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{A C}=\hat{i}-\hat{j}+3 \hat{k}$ such that $P$ is equidistant from the lines AB and AC. If $|\overrightarrow{\mathrm{AP}}|=\frac{\sqrt{5}}{2}$, then the area of the triangle ABP is:

Let $P Q R$ be a triangle such that $\overrightarrow{P Q}=-2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}}, a, b \in \mathbb{Z}$. Let S be the point on QR, which is equidistant from the lines PQ and PR. If $|\overrightarrow{\mathrm{PR}}|=9$ and $\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then the value of $3 a-4 b$ is $\_\_\_\_$

Let $\vec{a}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\vec{b}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ and $(\vec{a}-\vec{b}) \cdot \vec{c}=-97$, then $|\vec{c} \times \hat{\mathrm{k}}|^{2}$ is equal to

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda \hat{j}+2 \hat{k}, \lambda \in \boldsymbol{Z}$ be two vectors. Let $\vec{c}=\vec{a} \times \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $y z$-plane. If $|\overrightarrow{\mathrm{c}}|=\sqrt{53}$, then the maximum possible value of $(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}})^{2}$ 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 $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1,|\vec{b}|=4,|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is equal to

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

Let the foot of perpendicular from the point $(\lambda, 2, 3)$ on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$ and $\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}$ is equal to:

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 :

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\dfrac{x+3}{8} = \dfrac{y-4}{2} = \dfrac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

Let the image of the point $P(1, 6, a)$ in the line $L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$, $b > 0$, be $\left(\dfrac{a}{3}, 0, a+c\right)$. If $S(\alpha, \beta, \gamma)$, $\alpha > 0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

The square of the distance of the point $(-2, -8, 6)$ from the line $\dfrac{x-1}{1} = \dfrac{y-1}{2} = \dfrac{z}{-1}$ along the line $\dfrac{x+5}{1} = \dfrac{y+5}{-1} = \dfrac{z}{2}$ is equal to:

If $\left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}$, then the possible value(s) of $\alpha$ is (are)

If the point of intersection of the lines $\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{z+b+1}{7}$ and $\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{z-2a}{7}$ lies on $xy$-plane, then the value of $a + b$ is :

Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $y z-$ plane at the point Q. Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$.

Let $\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ be the image of the point $\mathrm{P}(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

Let the lines $\mathrm{L}_{1}: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $\mathrm{L}_{2}: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$, intersect at the point R. Let P and Q be the points lying on lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^{2}$ is equal to

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L, whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S}(a+b+c)$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$. Then the cosine of the acute angle between these lines is :