Question

What will be the perpendicular distance of a point P (7, 5 , -2) from the plane -2x +7y - 4z + 5 = 0 ?

• A. $\frac{23}{\sqrt{69}}$
• B. $\frac{34}{\sqrt{69}}$
• C. $\frac{57}{\sqrt{69}}$
• D. $\frac{62}{\sqrt{69}}$

Answer 1

The correct option is B

$\frac{34}{\sqrt{69}}$

We know that the perpendicular distance of a point $(x_1,y_1,z_1)$ from the plane ax+by+ cz+ d = 0 is $\frac{|(a{x}_1+b{y}_1+c{z}_1+d)|}{\sqrt{a^2+b^2+c^2}}$

We’ll do the appropriate substitutions

$\frac{|(-2.7+7.5-4.(-2)+5)|}{\sqrt{7^2+4^2+(-2{)}^2}}=\frac{34}{\sqrt{69}}$

So the distance from point P to the given plane will be $\frac{34}{\sqrt{69}}$