Question

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.

• A. $\left(12,-18,24\right)$
• B. $\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$
• C. $\left(\frac{12}{\sqrt{29}},-\frac{18}{\sqrt{29}},\frac{24}{\sqrt{29}}\right)$
• D. $(29,-29,29)$

Answer 1

The correct option is B $\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$

Let the coordinates of the foot of the perpendicular P from the origin to the plane be $\left(x_1,y_1,z_1\right)$
Then, the direction ratios of the line OP are $x_1,y_1,z_1$
Writing the equation of the plane in the normal
form, we have
$\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z-\frac{6}{\sqrt{29}}=0$
where
$\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}}$
are the direction cosines of OP.
Since D.Cs and direction ratios of a line are proportional, we have
$\frac{x_1}{\frac{2}{\sqrt{29}}}=\frac{y_1}{-\frac{3}{\sqrt{29}}}=\frac{z_1}{\frac{4}{\sqrt{29}}}=k$
$x_1=\frac{2k}{\sqrt{29}},y_1=-\frac{3k}{\sqrt{29}},z_1=\frac{4k}{\sqrt{29}}$
Substituting these in the equation of the plane, we get $k=\frac{6}{\sqrt{29}}$
Hence, the foot of the perpendicular is$\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$