Question

Find the length and the foot of the ${\perp }^r$ from the point $P(7,14,5)$ to the plane $2x+4y-z=2$. Also find the image of the point $P$ in the plane.

Answer 1

Let the foot of the perpendicular from the point onto the plane be $Q(x_1,y_1,z_1)$

The normal to the plane has direction ratios $(2,4,-1)$

Also, the direction ratios of the line joining the point $P(7,14,5)$ to the foot of perpendicular on the plane are given by $(7-x_1,14-y_1,5-z_1)$

The two ratios have to be proportional.

$\Rightarrow \frac{7-x_1}{x_1}=\frac{14-y_1}{y_1}=\frac{5-z_1}{z_1}=k$

$\Rightarrow x_1=\frac{7}{1+k},y_1=\frac{14}{1+k},z_1=\frac{5}{1+k}$

Substituting these points in the equation of plane, we have $2\times \frac{7}{1+k}+4\times \frac{14}{1+k}-\frac{5}{1+k}=2$

$\Rightarrow \frac{14+56-5}{1+k}=2$

$\Rightarrow k=\frac{63}{2}$

Thus, the foot of perpendicular $Q=(\frac{14}{65},\frac{28}{65},\frac{10}{65})$

Let point R be the image of point P about the given plane.

Q would be the midpoint of segment PR

Thus, $R=(\frac{28}{65}-7,\frac{56}{65}-14,\frac{20}{65}-5)$

$=(\frac{-427}{65},\frac{-854}{65},\frac{-305}{65})$