Question

If the equation of the plane which is at a distance of \(\dfrac{1}{3}\) 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. \(3\)
• B. \(\sqrt{5}\)
• C. \(\sqrt{13}\)
• D. \(2\)

Hint:

If a plane is perpendicular to a line, then the direction ratios of the line become the normal vector of the plane.

Answer 1

The Correct Option is A

Step 1: Use the normal direction.
The plane is perpendicular to a line with direction ratios $(1,2,2)$, so its normal is $(1,2,2)$ and the plane has the form $x + 2y + 2z + r = 0$.
Step 2: Match with the given form.
Comparing $x + py + qz + r = 0$ with $x + 2y + 2z + r = 0$, we read $p = 2$ and $q = 2$.
Step 3: Apply the distance from the origin.
The distance of $x+2y+2z+r=0$ from the origin is $\dfrac{|r|}{\sqrt{1^2+2^2+2^2}} = \dfrac{|r|}{3}$. This is given to be $\frac{1}{3}$.
Step 4: Solve for $r$.
So $\dfrac{|r|}{3} = \dfrac{1}{3}$, giving $|r| = 1$, i.e. $r = \pm 1$.
Step 5: Form the required sum of squares.
Then $p^2 + q^2 + r^2 = 2^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9$.
Step 6: Take the square root.
Therefore $\sqrt{p^2+q^2+r^2} = \sqrt9 = 3$.
\[ \boxed{3} \]