Question

A rod of length $2$ units whose one end is $(1,0,-1)$ and other end touches the plane $x-2y+2z+4=0$. Then

• A. the rod sweeps a figure whose volume is $\pi$ cubic units.
• B. the area of the figure which the rod traces on the plane is $2\pi$ units.
• C. the length of projection of the rod on the plane is $\sqrt{3}$ units.
• D. the centre of the region which the rod traces on the plane is $(\frac{2}{3},\frac{2}{3},\frac{-5}{3})$

Answer 1

Answer: (A), (C), (D)

the centre of the region which the rod traces on the plane is $(\frac{2}{3},\frac{2}{3},\frac{-5}{3})$


Let $AB$ be the rod.
Given $AB=2$
So the rod sweeps out a cone.
Perpendicular distance of $A(1,0,-1)$ from the plane
$x-2y+2z+4=0$ is
$AC=\frac{|1-2+4|}{\sqrt{1+4+4}}=1$
$\therefore$ Radius of the base of the cone, $BC=\sqrt{2^2-1^2}=\sqrt{3}$

Hence, the volume of the cone
$=\frac{1}{3}\pi (\sqrt{3}{)}^2\cdot 1=\pi$

Area of the circle which the rod traces on the plane is
$\pi (\sqrt{3}{)}^2=3\pi$.

Let centre of the circle be $C(x,y,z)$
Then equation of the line joining points $A$ and $C$,
$\frac{x-1}{1}=\frac{y-0}{-2}=\frac{z+1}{2}=k\Rightarrow x=k+1,y=-2k,z=2k-1$
Putting this in the equation of plane,
$k+1+4k+4k-2+4=0\Rightarrow k=-\frac{1}{3}$
$\Rightarrow x=\frac{2}{3},y=\frac{2}{3},z=\frac{-5}{3}$