If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\overrightarrow{\mathrm{r}}=\lambda(\hat{i}+\mathrm{a} \hat{j}+\mathrm{b} \hat{k}) \quad$ and $...

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1$. Then $\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$ is :

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 points on the lines...

Consider the lines $\mathrm{L}_1: x-1=y-2=\mathrm{z}$ and $\mathrm{L}_2: x-2=y=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be...

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

The line $L_1$ is parallel to vector $\vec{a}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and passes through the point $(7,6,2)$ and the line $L_2$ is parallel to vector...

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$, be...

If the image of the point $\mathrm{P}(1,0,3)$ in the line joining the points $\mathrm{A}(4,7,1)$ and $\mathrm{B}(3,5,3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to...

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$, then $\left|72-|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2\right|$ is equal to

Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2 y+z-2=0$ be $P$. If the distance of the point $Q(6-2, \alpha), \alpha>Q$...