Question

Foot of perpendicular drawn from the origin to the plane $2x-3y+4z=29$ is _____

• A. $(5,-1,4)$
• B. $(2,-3,4)$
• C. $(7,-1,3)$
• D. $(5,-2,3)$

Answer 1

The correct option is B $(2,-3,4)$

Given equation of the plane is $2x-3y+4z=\mathrm{29........}\left(1\right)$

Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $(x_1,y_1,z_1)$.

The direction cosines of the plane are $(2,-3,4)$.

Therefore, $\sqrt{2^2+(-3{)}^2+4^2}=\sqrt{4+9+16}=\sqrt{29}$

Divide eqn 1 on both sides by $\sqrt{29}$

$\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{29}{\sqrt{29}}$

Therefore,

$(\frac{2}{\sqrt{29}}\times \frac{29}{\sqrt{29}}),(-\frac{3}{\sqrt{29}}\times \frac{29}{\sqrt{29}}),(\frac{4}{\sqrt{29}}\times \frac{29}{\sqrt{29}})=(2,-3,4)$

Hence, the coordinates of the foot of the perpendicular are $(2,-3,4)$.