Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 4: EXERCISE-4

Find the angle between the two straight lines whose direction cosines $l,m,n$ are given by $2l+2m-n=0$ and $mn+nl+lm=0$

A variable plane is at a constant distance $p$ form the origin and meet the coordinate axes in points $A,B$ and $C$ respectively. through these points, planes are drawn parallel to the coordinates planes. Find the locus of their point of intersection 

$P$ is any point on the plane $lx+my+nz=p$. $A$ point $Q$ taken on the $OP$ (where $O$ is the origin) such that $OP·OQ=p^2$. Show that the locus of $Q$ is $p\left(lx+my+nz\right)=x^2+y^2+z^2$

Find the point where the line of intersection of the planes $x-2y+z=1$ and $x+2y-2z=5$, intersects the plane $2x+2y+z+6=0$

Find the foot and hence the length of the perpendicular from the point $(5,7,3)$ to the line $\frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}$. Also find the equation of the plane in which the perpendicular and the given straight line lie.

Find the equation of the plane containing the line $\frac{x-1}{2}=\frac{y}{3}=\frac{z}{2}$ and parallel to the line $\frac{x-3}{2}=\frac{y}{5}=\frac{z-2}{4}$. Find also the S.D. between two lines.

Find the equations to the line which can be drawn from the point $(2,-1,3)$ perpendicular to the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{4}=\frac{y}{5}=\frac{z+3}{3}$

Find the equations to the line of greatest slope through the point $(7,2,-1)$ in the plane $x-2y+3 z=0$ assuming that the axes are so placed that the plane $2x+3y-4z=0$ is horizontal.