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

Let $P\left(1,1,1\right), Q\left(2,3,5\right), R\left(-1,0,2\right)$ be three points, then equation of a plane parallel to the plane $PQR$ which is at a distance of $5$ units from origin is

If $\theta$ is the angle between the lines in which the planes $3x-7y-5z=1$ and $5x-13y+3z+2=0$ cuts the plane $8x-11y+2z=0,$ then $\sin \theta$ is

The plane $4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane $5x+3y+10z=25$. The equation of the plane in its new position is $x-4y+6z=k$ where $k$ is

A line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$, respectively. If length $PQ=d$, then $d^2$ is equal to 

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of the two lines inclined to each other at an angle $\theta$, then the direction cosines of the bisector of the angle between these lines are-

The equation of a plane bisecting the angle between the plane $2x-y+2z+3=0$ and $3x-2y+6z+8=0$ is-

The line $\frac{x-4}{k}=\frac{y-2}{1}=\frac{z-k^2}{2}$ lies in the plane $2 x-4 y+z=1.$ Then the value of $k$ cannot be

Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $x+y+z=0, lx-nz=0$ and $x-2y+z=0$ equal to $9$ units. If the locus of the point $P$ is $x^2+y^2+z^2=9,$ then the value of $l-n$ is equal to