Vectors & 3D Geometry PYQ — Page 8

Let $\mathrm{P}$ be the point of intersection of the lines $\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$ and $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$. Then, the shortest distance of P from the line $4 x=2 y=z$ is

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 $P\text{and}Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha ,\beta ,\gamma ),$ then ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}$ 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 $Q$ and $R$ be the feet of perpendiculars from the point $P(a,a,a)$ on the lines $x=y,z=1$ and $x=-y,z=-1$ respectively. If $\angle QPR$ is a right angle, then $12{a}^{2}$ is equal to ________

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

Let ${L}_{1}:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda (\hat{i}-\hat{j}+2\hat{k}),\lambda \in R$, ${L}_{2}:\vec{r}=(\hat{j}-\hat{k})+\mu (3\hat{i}+\hat{j}+p\hat{k}),\mu \in R$ and ${L}_{3}:\vec{r}=\delta (l\hat{i}+m\hat{j}+n\hat{k}),\delta \in R$ be three lines such that ${L}_{1}$ is perpendicular to ${L}_{2}$ and ${L}_{3}$ is perpendicular to both ${L}_{1}$ and ${L}_{2}$. Then the point which lies on ${L}_{3}$ is

If ${d}_{1}$ is the shortest distance between the lines $x+1=2y=-12z,x=y+2=6z-6$ and ${d}_{2}$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5},\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32\sqrt{3}{d}_{1}}{{d}_{2}}$ is :

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 $P(3,2,3),Q(4,6,2)$ and $R(7,3,2)$ be the vertices of $\Delta \mathrm{PQR}$. Then, the angle $\angle \mathrm{QPR}$ is

Let the image of the point $(1,0,7)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ be the point $(\alpha ,\beta ,\gamma )$. Then which one of the following points lies on the line passing through $(\alpha ,\beta ,\gamma )$ and making angles $\frac{2\pi }{3}$ and $\frac{3\pi }{4}$ with $y-$axis and $z-$axis respectively and an acute angle with $x-$axis?

The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14{l}^{2}$ is equal to _____.

The position vectors of the vertices $A,B$ and $C$ of a triangle are $2\hat{i}-3\hat{j}+3\hat{k},2\hat{i}+2\hat{j}+3\hat{k}$ and $-\hat{i}+\hat{j}+3\hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $\mathrm{AD}$ of $\angle \mathrm{BAC}$ where $D$ is on the line segment $\mathrm{BC}$, then $2{l}^{2}$ equals :

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

Let the position vectors of the vertices $A,B$ and $C$ of a triangle be $2\hat{i}+2\hat{j}+\hat{k},\hat{i}+2\hat{j}+2\hat{k}$ and $2\hat{i}+\hat{j}+2\hat{k}$ respectively. Let ${l}_{1},{l}_{2}$ and ${l}_{3}$ be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides $\mathrm{AB},\mathrm{BC}$ and $\mathrm{CA}$ respectively, then ${l}_{1}^{2}+{l}_{2}^{2}+{l}_{3}^{2}$ equals :

The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2\hat{j}+2\hat{k}$ and $\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}$ is acute, is _____.

If the shortest distance between the lines $\frac{x-\lambda }{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1$, then the sum of all possible values of $\lambda$ is

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________

The shortest distance between lines ${L}_{1}$ and ${L}_{2}$, where ${L}_{1}:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and ${L}_{2}$ is the line passing through the points $A(-4,4,3),B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is

Between the following two statements: Statement I : Let $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r}=\vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r}=0$ is of magnitude $\sqrt{10}$. Statement II : In a triangle $A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}$.

A line with direction ratio $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2y=2z$ respectively at the point $P$ and $Q$. if the length of the perpendicular from the point $(1,2,12)$ to the line $\mathrm{PQ}$ is $l$, then ${l}^{2}$ is

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

If the shortest distance between the lines $\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R} \end{array}$ is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals

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___________