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

The shortest distance of the point $\left(1,3,5\right)$ from $x^2+y^2=0$ is

If the extremities of the diagonal of a square are $\left(1,-2,3\right)$ and $\left(3,-4,3\right)$, then the length of the side is

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

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 ${\sin }^{-1}\left(\sin \lambda \right)+{\cos }^{-1}\left(\cos \lambda \right)=$

If the lines $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-\lambda }{3}$ and $\frac{x}{1}=\frac{y+2}{2}=\frac{z}{4}$ intersect each other, then $\lambda$ lies in the interval

The reflection point of the point $\left(0,3,-2\right)$ in the line $\frac{1-x}{2}=2-y=z+1$ is

The plane through the intersection of planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to $Y-$axis, also passes through the point

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