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Q.
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 :
Three Dimensional Geometry
Solution:
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}} $
