Question

The perpendicular distance form point $(2,-3,6)$ to plane $3x-6y+2z+10=0$ is _________.

• A. $-\frac{46}{7}$
• B. $\frac{13}{7}$
• C. $\frac{46}{7}$
• D. $\frac{10}{7}$

Answer 1

The correct option is C $\frac{46}{7}$

From the given equation of the plane
we have $a=3$, $b=-6$, $c=2$, $d=-10$ and given point $(x_1,y_1,z_1)=(2,-3,6)$
$\therefore$ Perpendicular distance $p=$
$=\left|\frac{a{x}_1+b{y}_1+x{z}_1-d}{\sqrt{1^2+b^2+c^2}}\right|$
$=\left|\frac{3\left(2\right)+(-6)(-3)+2\left(6\right)-(-10)}{\sqrt{9+36+4}}\right|$
$=\left|\frac{6+18+12+10}{\sqrt{9+36+4}}\right|$
$\therefore p=\frac{46}{\sqrt{49}}$
$\therefore$ perpendicular distance $=\frac{46}{7}$.