• # question_answer Let Q be the foot of perpendicular from the origin to the plane $4x-3y+z+13=0$ and R be point (-1, 1, -6) on the plane Then length QR is :     JEE Main  Online Paper (Held On 22 April 2013) A)  $\sqrt{14}$                         B)  $\sqrt{\frac{19}{2}}$ C)  $3\sqrt{\frac{7}{2}}$                         D)  $\frac{3}{\sqrt{2}}$

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 $4,-3,\text{ }1.$ 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 $(4r,-3r,r)$ 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 $4x-3y+z+13=0$ $\therefore$  $8r+\frac{9}{2}+r+\frac{r}{2}+13=0$ $\Rightarrow$               $r=\frac{-13}{8+\frac{9}{2}+\frac{1}{2}}=\frac{-26}{26}=-1$ $\therefore$  $Q=\left( -2,\frac{3}{2},-\frac{1}{2} \right)$ $QR=\sqrt{{{(-1+2)}^{2}}+{{\left( 1-\frac{3}{2} \right)}^{2}}+{{\left( -6+\frac{1}{2} \right)}^{2}}}$ $=\sqrt{1+\frac{1}{4}+\frac{121}{4}}=3\sqrt{\frac{7}{2}}$