Question

If the equation of the plane which is at a distance of $\dfrac{1}{\sqrt{3}}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x + py + qz + r = 0$, then $\sqrt{p^2 + q^2 + r^2} =$

• A. $\sqrt{3}$
• B. $\sqrt{5}$
• C. $\sqrt{13}$
• D. $2$

Hint:

Use distance-from-origin formula for plane and convert direction ratios to normal form.

Answer 1

The Correct Option is A

Distance from origin to plane $Ax + By + Cz + D = 0$ is $\dfrac{|D|}{\sqrt{A^2 + B^2 + C^2}}$
Given: $x + py + qz + r = 0$ with direction ratios $(1, 2, 2)$ ⇒ normal vector
So normal vector has magnitude = $\sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$
Distance = $\dfrac{|r|}{\sqrt{1^2 + 2^2 + 2^2}} = \dfrac{|r|}{3} = \dfrac{1}{\sqrt{3}} \Rightarrow |r| = \dfrac{3}{\sqrt{3}} = \sqrt{3}$
So $\sqrt{p^2 + q^2 + r^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$