Question

Find the perpendicular distance to the plane $3x+2y-5z-13=0$ from the point $(5,3,4)$.

Answer 1

Perpendicular distance from $P(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is

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

Perpendicular distance from $P(5,3,4)$ to the plane $3x+2y-5z-13=0$ is

$\frac{|3\times 5+2\times 3-5\times 4-13|}{\sqrt{{\left(3\right)}^2+{\left(2\right)}^2+{(-5)}^2}}$

$=\frac{|15+6-20-13|}{\sqrt{9+4+25}}$

$=\frac{14}{\sqrt{38}}$

$=\frac{14}{\sqrt{38}}\times \frac{\sqrt{38}}{\sqrt{38}}$

$\therefore$ perpendicular distance$=\frac{7\sqrt{38}}{19}$ units.