Vectors & 3D Geometry PYQ — Page 8

The image of the point (1,2,3) in the plane x+y+z=9 is:

Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $\mathrm{P}$, be $(\alpha, \beta, \gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{c}}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be anit vector along $\vec{b}+\vec{c}$. If $\vec{r} \cdot \vec{a}=3$, then $3 \lambda$ is equal to:

Let a line passing through the point $(-1,2,3)$ intersect the lines ${L}_{1}:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$ at $M(\alpha ,\beta ,\gamma )$ and ${L}_{2}:\frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$ at $N(a,b,c)$. Then the value of $\frac{(\alpha +\beta +\gamma {)}^{2}}{(a+b+c{)}^{2}}$ equals ________________.

The projection of vector a=2i-3j+6k on vector b=i+2j+2k is:

A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $P(a,b,c)$ where $a,b,c$ are non-negative integers, on the line $AB$ lies at a distance of $21$ units, from the point $A$. The distance between the points $P(a,b,c)$ and $Q(4,-12,3)$ is equal to ______.

Let $O$ be the origin and the position vector of $A$ and $B$ be $2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot(\vec{b}-\vec{c})=0$, then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to___________

If the line $\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7}$, then $4 \lambda+9 \mu$ is equal to :

Let $\mathrm{d}$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $\mathrm{d}^2+6$ is equal to :

Let $O$ be the origin, and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\vec{OM}\cdot \vec{ON}$ is equal to _________.

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to___________

Let $PQR$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $PQ$. The distance of the centroid of $\Delta PQR$ from the point of intersection of the line $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is

The distance of the point $Q(0,2,–2)$ form the line passing through the point $P(5,–4,3)$ and perpendicular to the lines $\vec{r}=(-3\hat{i}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+5\hat{k}),\lambda \in \mathbb{R}$ and $\vec{r}=(\hat{i}-2\hat{j}+\hat{k})+\mu (-\hat{i}+3\hat{j}+2\hat{k}),\mu \in \mathbb{R}$ is

The square of the distance of the image of the point $(6,1,5)$ in the line $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$, from the origin is _________

Let $\vec{\mathrm{OA}}=\vec{a},\vec{\mathrm{OB}}=12\vec{a}+4\vec{b}$ and $\vec{\mathrm{OC}}=\vec{b}$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $\mathrm{OA}$ and $\mathrm{OC}$, then $\frac{\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}\mathrm{OABC}}{\mathrm{area}\mathrm{of}S}$ is equal to _____

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1,|\vec{b}|=4$ and $\vec{a}\cdot \vec{b}=2$. If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b}$ and the angle between $\vec{b}$ and$\vec{c}$ is $\alpha$, then $192{\mathrm{sin}}^{2}\alpha$ is equal to _________

Let $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k},\alpha ,\beta \in R$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi }{4}$ and ${|\vec{b}|}^{2}=6$, If $\vec{a}\cdot \vec{b}=3\sqrt{2}$, then the value of $({\alpha }^{2}+{\beta }^{2})|\vec{a}\times \vec{b}{|}^{2}$ is equal to

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+5 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$. If $\vec{a} \cdot \vec{c}=-29$, then $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

Let a unit vector $\hat{u}=x\hat{i}+y\hat{j}+z\hat{k}$ make angles $\frac{\pi }{2},\frac{\pi }{3}$ and $\frac{2\pi }{3}$ with the vectors $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k},\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$ and $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$ respectively. If $\vec{v}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{\sqrt{2}}\hat{k}$, then \(|\hat{u}-\vec{v}|^2\) is equal to

Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. If $\vec{a} \cdot \vec{c}=130$, then $\vec{b} \cdot \vec{c}$ is equal to _______

Let $\vec{a}=3\hat{i}+\hat{j}-2\hat{k},\vec{b}=4\hat{i}+\hat{j}+7\hat{k}$ and $\vec{c}=\hat{i}-3\hat{j}+4\hat{k}$ be three vectors. If a vectors $\vec{p}$ satisfies $\vec{p}\times \vec{b}=\vec{c}\times \vec{b}$ and $\vec{p}\cdot \vec{a}=0$, then $\vec{p}\cdot (\hat{i}-\hat{j}-\hat{k})$ is equal to

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :