Vectors & 3D Geometry PYQ

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)

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 :

Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of L from the point $(-2, \mathrm{r}, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of r is :

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:

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 $\_\_\_\_$。

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 :

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

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 :

The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$

Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ 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 :

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{c} \cdot (\vec{a} - 2\vec{b})$ is equal to _______.

Let $\vec{a}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{b}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{c}=\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\vec{v}=\vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9 p^{2}$ is equal to

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 $\overrightarrow{\mathrm{a}}=-\hat{i}+\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\hat{i}-\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to :

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

The shortest distance between the lines $\vec{r}=\left(\dfrac{1}{3}\hat{i}+2\hat{j}+\dfrac{8}{3}\hat{k}\right)+\lambda(2\hat{i}-5\hat{j}+6\hat{k})$ and $\vec{r}=\left(-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{k}\right)+\mu(\hat{j}-\hat{k})$, $\lambda,\mu \in \mathbb{R}$, is:

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2 \vec{a}+k \vec{b}+k \vec{c}|=3$, the positive value of k is

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}$. If $\vec{a}\cdot\vec{c} = 15$, then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$ is equal to:

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 :

Let a line $L$ be perpendicular to both the lines $L_1: \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7}$ and $L_2: \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \dfrac{x - \dfrac{8}{7}}{2} = \dfrac{y - \dfrac{4}{7}}{1} = \dfrac{z}{2}$, then $\tan\theta$ is equal to:

The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is