Question

Let Q be the foot of perpendicular from the origin to the plane $4x-3y+z+13=0$ and R be a point ${-1,1,6}$ 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

The Correct Option is 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 \(4x - 3y + z + 13 = 0\)

\(\therefore 8r + \frac{9}{2} r + \frac{r}{2} +13 =0\) \(\Rightarrow r = \frac{-13}{8+ \frac{9}{2} + \frac{1}{2}}= \frac{-26}{26} = - 1\)

\(\therefore Q = \left(-2 , \frac{3}{2} , - \frac{1}{2}\right)\) \(QR = \sqrt{\left(-1+2\right)^{2} + \left(1- \frac{3}{2}\right)^{2} + \left(-6+ \frac{1}{2}\right)^{2}}\) \(= \sqrt{1+\frac{1}{4} + \frac{121}{4}} = 3 \sqrt{\frac{7}{2}}\)

so, The correct option is(C):  \(3{\sqrt\frac{7}{2}}\).