Question

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

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

Answer 1

The correct option is D

$(2,-3,4)$

Finding the foot of the plane:

Given equation of plane $2-3y+4z=29$

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

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

Therefore,

$\begin{array}{rcl}D& =& \sqrt{l^2+m^2+n^2}\\ & =& \sqrt{2^2+{(-3)}^2+4^2}[l=2,m=-3,n=4]\\ & =& \sqrt{29}\end{array}$

Divide equation (i) on both side by$\sqrt{29}$

Therefore,

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

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

Hence, the correct option is (D).