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

Answer: (C)

$3 \sqrt{\frac{7}{2}}$

Answer 2

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): $3{\sqrt\frac{7}{2}}$.