Question

A plane $P=0$ is the perpendicular bisector of the line joining the points $\left(2,3,4\right)$ and $\left(6,7,8\right).$ The perpendicular distance of $P=0$ from the origin is

• A. $4\sqrt{3}$ units
• B. $5\sqrt{3}$ units
• C. $6\sqrt{3}$ units
• D. $8\sqrt{3}$ units

Answer 1

Answer: (B)

From diagram mid-point of $AB$ must lie on the plane
$M\left(\frac{2+6}{2},\frac{3+7}{2},\frac{4+8}{2}\right)\equiv \left(4,5,6\right)$
$AB$ is perpendicular to the required plane
Direction ratios of $\overrightarrow{\mathrm{AB}}=<4,4,4>=<1,1,1>$
Equation of the required plane is
$1\left(x-4\right)+1\left(y-5\right)+1\left(z-6\right)=0$
$\Rightarrow x+y+z=15$
The perpendicular distance from the origin to the plane $=\left|\frac{0+0+0-15}{\sqrt{3}}\right|=5\sqrt{3}$ units