NCERT Solutions for Chapter: Three Dimensional Geometry, Exercise 1: EXERCISE

Find the position vector of a point $A$ in space such that $\overrightarrow{OA}$ is inclined at $60°$ to $OX$ and at $45°$ to $OY$ and $\left|\overrightarrow{OA}\right|=10\text{ units.}$

Prove that the lines $x=py+q, z=ry+s$ and $x=p^{\text{'}}y+q^{\text{'}}, z=r^{\text{'}}y+s^{\text{'}}$ are perpendicular if $p{p}^{\text{'}}+r{r}^{\text{'}}+1=0$.

 Find the equation of the plane through the points $\left(2,1,-1\right)\text{ and }\left(-1,3,4\right)$ and perpendicular to the plane $x-2y+4z=10$.

Find the shortest distance(in units) between the lines given by

$\overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}\text{ and }\overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})$.

Find the equation of the plane through the intersection of the planes $\overrightarrow{r}·(\hat{i}+3\hat{j})-6=0$ and $\overrightarrow{r}·(3\hat{i}-\hat{j}-4\hat{k})=0$, whose perpendicular distance from origin is unity.

$\overrightarrow{AB}\mathit{=}3\hat{i}\mathit{-}\hat{j}\mathit{+}\hat{k}$ and $\overrightarrow{CD}\mathit{=}\mathit{-}3\hat{i}+2\hat{j}+4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i}+7\hat{j}+4\hat{k}$ and $-9\hat{j}+2\hat{k}$, respectively. Find the position vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{AB}$ and $\overrightarrow{CD}$ both.

The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is 

The vector equation of the line through the points $(3, 4, -7)$ and $(1, -1, 6)$ is