Question

The minimum distance between the origin and the plane which is perpendicular bisector of the line joining the points $(1,3,5)$ and $(3,7,-1)$, is

• A. $\frac{6}{\sqrt{14}}$
• B. $\frac{3}{7}$
• C. $\frac{3}{\sqrt{14}}$
• D. $\frac{6}{7}$

Answer 1

The correct option is A $\frac{6}{\sqrt{14}}$

Vector joining the points $(1,3,5)$ and $(3,7,-1)$ is $2\hat{i}+4\hat{j}-6\hat{k}$
So, the normal vector to the plane is, $\overrightarrow{n}=\hat{i}+2\hat{j}-3\hat{k}$
Equation of the plane which is perpendicular bisector of the line joining the two points, is $x+2y-3z=d$
Midpoint of $(1,3,5)$ and $(3,7,-1)$ is, $(2,5,2)$.
As the mid-point lies on the plane,
so, $2+2\times 5-3\times 2=d$
$\Rightarrow d=6$
Therefore, the equation of the plane is,
$x+2y-3z-6=0$

The minimum distance from the origin is the perpendicular distance, $\left|\frac{-6}{\sqrt{1^2+2^2+(-3{)}^2}}\right|=\frac{6}{\sqrt{14}}$