Question

The distance of the point (2, 1, – 2) from the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-3}$ measured parallel to the plane x + 2y + z = 4 is

• A. $\sqrt{10}$
• B. $\sqrt{20}$
• C. $\sqrt{5}$
• D. $\sqrt{30}$

Answer 1

The correct option is D $\sqrt{30}$

Let direction ratios of PQ and a, b, c
$\frac{x+1}{-1}=\frac{y}{2}=\frac{z-1}{3}$

Then, $\frac{x-2}{a}=\frac{y-1}{b}=\frac{z+2}{c}$
(Let PQ = r)
$Q\equiv$ (ar + 2, br + 1, cr – 2)
Which lie on $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-3}{-3},then\frac{ar+2-1}{2}=\frac{br+1+1}{2}=\frac{cr-2-3}{-3}\Rightarrow \frac{ar+1}{2}=\frac{br+2}{2}=\frac{cr-5}{-3}=\lambda \left(say\right)\therefore a=\frac{2\lambda -1}{r},b=\frac{2\lambda -2}{r},c=\frac{-3\lambda +5}{r}$
Given, PQ is parallel to x + 2y +z = 4, then
a + 2b + c = 0
$or\frac{2\lambda -1}{r}+\frac{4\lambda -4}{r}+\frac{-3\lambda +5}{r}=0\Rightarrow 3\lambda =0\Rightarrow \lambda =0thena=-\frac{1}{r},b=-\frac{2}{r},c=-\frac{5}{r}\therefore Q=(1,–1,3)\therefore PQ\equiv \sqrt{(2-1{)}^2+(1+1{)}^2+(3+2{)}^2}=\sqrt{30}$