Vectors & 3D Geometry PYQ — Page 5

The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$ . Then, which of the following points lies on the line L ?

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is

Let P be the image of the point $\mathrm{Q}(7,-2,5)$ in the line $\mathrm{L}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is ________.

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

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive y - and z -axes, respectively, is half of the angle that this line makes with the positive x -axes. Then the sum of all possible values of the angle $\beta$ is

Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and ( $0,0,1$ ). Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements (S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and (S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}$ and $\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :

Let the point A divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $\mathrm{r}: 1(\mathrm{r}\gt0)$. If O is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10$, then the value of r is :

Let $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\mathrm{L}_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$ ?

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

Let \(\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}\) and \(\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\) be two lines. Let \(L_3\) be a line passing through the point \((\alpha, \beta, \gamma)\) and be perpendicular to both \(L_1\) and \(L_2\). If \(L_3\) intersects \(\mathrm{L}_1\), then \(|5 \alpha-11 \beta-8 \gamma|\) equals :

Let $A$ be the point of intersection of the lines $L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let $B$ and $C$ be the point on the lines $L_1$ and $L_2$ respectively such that $\mathrm{AB}=\mathrm{AC}=\sqrt{15}$. Then the square of the area of the triangle ABC is :

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \vec{a}+2 \vec{b}$ and $3 \vec{a}-\lambda \vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1,3]$ is :

Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b}$, $(a \lt b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is :-

Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $A B C D$ be $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $A B C$ meets the median line segment through $A$ of the triangle $A B C$ at the point $E$. If the length of $A D$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6 \sqrt{2}}$, then the position vector of $E$ is

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Let the vertices $Q$ and $R$ of the triangle $P Q R$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, Q R=5$ and the coordinates of the point $P$ be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :

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 $A B C, \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2}$.

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 _____.

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 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 :

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