Question

Let Q be the foot of perpendicular from the origin to the plane 4x - 3y + z + 13 = 0 and R be a point (- 1, - 6) on the plane. Then length QR is :

• A. $\sqrt{14}$
• B. $\sqrt{\frac{19}{2}}$
• C. $3\sqrt{\frac{7}{2}}$
• D. $\frac{3}{\sqrt{2}}$

Answer 1

Answer: (C)

Let P be the image of O in the given plane.

Equation of the plane, 4x - 3y + z + 13 = 0
OP is normal to the plane, therefore direction ratio of OP are proportional to 4, - 3, 1
Since OP passes through (0, 0, 0) and has direction ratio proportional to 4, -3, 1. Therefore equation of OP is
$\frac{x-0}{4}=\frac{y-0}{-3}=\frac{z-0}{1}=r(let)$
$\therefore x=4r,y=-3r,z=r$
Let the coordinate of P be $\left(4r,-3r,r\right)$
Since Q be the mid point of OP
$\therefore Q=\left(2r,-\frac{3}{2}r,\frac{r}{2}\right)$
Since Q lies in the given plane
$4x-3y+z+13=0$
$\therefore 8r+\frac{9}{2}r+\frac{r}{2}+13=0$
$\Rightarrow r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1$
$\therefore Q=\left(-2,\frac{3}{2},-\frac{1}{2}\right)$
$QR=\sqrt{{\left(-1+2\right)}^2+{\left(1-\frac{3}{2}\right)}^2+{\left(-6+\frac{1}{2}\right)}^2}$
$=\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3\sqrt{\frac{7}{2}}$