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: Identify the normal vector of the plane.
A plane perpendicular to a line has a normal vector parallel to the direction ratios of the line.
Given direction ratios are
\[ (1,2,2) \] Hence, the normal vector of the plane is
\[ \vec{n}=(1,2,2) \]

Step 2: Write the equation of the plane.
The general equation of the plane is given as
\[ x+py+qz+r=0 \] Comparing with the normal vector \((1,2,2)\), we get
\[ p=2,\qquad q=2 \] So, the plane becomes
\[ x+2y+2z+r=0 \]

Step 3: Use the distance formula of a plane from the origin.
Distance of the plane
\[ ax+by+cz+d=0 \] from the origin is
\[ \frac{|d|}{\sqrt{a^2+b^2+c^2}} \] Here,
\[ a=1,\quad b=2,\quad c=2,\quad d=r \] Given distance is
\[ \frac{1}{3} \] Therefore,
\[ \frac{|r|}{\sqrt{1^2+2^2+2^2}}=\frac{1}{3} \] \[ \frac{|r|}{\sqrt{1+4+4}}=\frac{1}{3} \] \[ \frac{|r|}{3}=\frac{1}{3} \] \[ |r|=1 \] Thus,
\[ r=\pm 1 \]

Step 4: Calculate \(\sqrt{p^2+q^2+r^2}\).
We have
\[ p=2,\qquad q=2,\qquad r^2=1 \] Hence,
\[ \sqrt{p^2+q^2+r^2} = \sqrt{2^2+2^2+1} \] \[ = \sqrt{4+4+1} \] \[ = \sqrt{9} \] \[ =3 \]

Step 5: Final conclusion.
Therefore,
\[ \boxed{3} \]