Question

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

• A. the rod sweeps a solid structure whose volume is $\pi$ cubic units.
• B. the area of the region which the rod traces on the plane is $2\pi$ square 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 $(\frac{2}{3},\frac{2}{3},-\frac{5}{3})$.


The rod sweeps out the figure which is a cone.
Perpendicular distance of point $A(1,0,-1)$ from the plane is $\frac{|1-0-2+4|}{\sqrt{9}}=1$ unit.
Slant height of the cone is $2$ units.
Then, the radius of the base of the cone is $\sqrt{4-1}=\sqrt{3}$
Hence, volume of the cone is $\frac{\pi }{3}(\sqrt{3}{)}^2\cdot 1=\pi$ cubic units.

Area of the circle on the plane which the rod traces is $3\pi .$

If the centre of the circle is $Q(x,y,z),$ then
$\frac{x-1}{1}=\frac{y-0}{-2}=\frac{z+1}{2}=\frac{-(1-0-2+4)}{1^2+(-2{)}^2+2^2}$
So, $Q(x,y,z)=(\frac{2}{3},\frac{2}{3},\frac{-5}{3})$