Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Level 1

$A(-1,2,-3), B(5,0,-6), C(0,4,-1)$ are the vertices of a triangle $ABC$. The direction cosines of internal bisector of $\angle BAC$ are

The point of intersection of the lines $\frac{x-6}{-6}=\frac{y+4}{4}=\frac{z-4}{-8}$ and $\frac{x+1}{2}=\frac{y+2}{4}=\frac{z+3}{-2}$ is

The value of $p$, so that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{p}$ and $\frac{x}{p}=\frac{y}{2}=\frac{z}{3}$ are at right angles, is

The angle between the lines $6x=3y=4z$ and $2x=-y=z$ is

The perpendicular distance of the point $\left(2,4,-1\right)$ from the line$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ is

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda }$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar, then ${\sin }^{-1}\left(\sin \lambda \right)+{\cos }^{-1}\left(\cos \lambda \right)=$

The two lines $L_1 : \overrightarrow{r}=(\hat{i}+5\hat{j}+5\hat{k})+t(4\hat{i}-4\hat{j}+5\hat{k})$ and $L_2 : \overrightarrow{r}=(2\hat{i}+4\hat{j}+5\hat{k})+s(8\hat{i}-3\hat{j}+\hat{k})$ are such that

If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar, then the value of $k$ is _______.