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# Find the $\perp$ distance of the point (2,3,-5) from the plane $x+2y-2z-9=0$

Toolbox:
• Distance of any point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by $\bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
Given point $Q(2,3,-5)$ and Eqn. of plane $x+2y-2z-9=0$
We know that the distance of any point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by
$\bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
$\perp$ distance =$\bigg|\large\frac{2+6+10-9}{\sqrt {1^2+2^2+(-2)^2}}\bigg|=\frac{9}{3}$
$=3$

edited May 23, 2013