Question

The perpendicular distance of a point P (7,5 ,-2) from the plane -2x +7y -4z + 5 = 0 is units.

• 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 $d=\frac{|(a{x}_1+b{y}_1+c{z}_1+d)|}{\sqrt{a^2+b^2+c^2}}$

Making the appropriate substitutions,

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