Question

The foot of perpendicular form the point $(7,14,5)$ to the plane $2x+4y-z=2$, is given by

• A. $(1,8,2)$
• B. $(1,2,8)$
• C. $(1,-2,8)$
• D. $(1,2,-8)$

Answer 1

Answer: (B)

$(1,2,8)$

Point $(7,14,5)$

Plane:$2x+4y-z=2$

Directions of normal to plane $(2,4,-1)$

Equation of line perpendicular to plane & passing through $(7,14,5)$

$\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=k$

Let $(2k+7,4k+14,-k+5)$ be the foot of $\perp$.

$\therefore$ It should satisfy plane.

$2(2k+7)+4(4k+14)-(-k+5)=2$

$\Rightarrow k=-3$

$\therefore$ foot of $\perp =(1,2,8)$