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 :

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 $\overrightarrow{\mathrm{a}}=-\hat{i}+2 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=8 \hat{i}+7 \hat{j}-3 \hat{k}$ 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 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 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 $\mathrm{P}(\alpha, \beta, \gamma)$ 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

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \dfrac{3}{4}, \dfrac{-1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to:

Let a line $L_1$ pass through the origin and be perpendicular to the lines $L_2: \vec{r} = (3+t)\hat{i} + (2t-1)\hat{j} + (2t+4)\hat{k}$ and $L_3: \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$, $t, s \in \mathbb{R}$. If $(a, b, c)$, $a \in \mathbb{Z}$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to ________.

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 $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ 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:

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:

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 image of the point $P(0, -5, 0)$ in the line $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ be the point $R$ and the image of the point $Q\left(0, \dfrac{-1}{2}, 0\right)$ in the line $\dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is __________.

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(a\hat{i} - \hat{j})$, $a \neq 0$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + a\hat{k})$ from the origin is:

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

If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P, then $a+b$ is equal to $\_\_\_\_$.

A line with direction ratios $1, -1, 2$ intersects the lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z+1}{3}$ and $\dfrac{x+1}{-1} = \dfrac{y-2}{1} = \dfrac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is :

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:

Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to