# The distance of the plane $2x-3y+6z+14=0$ from origin is ?

$\begin{array}{1 1} 2 \\ 3 \\ 11 \\ 14 \end{array}$

## 1 Answer

Toolbox:
• The distance of the plane $ax+by+cz+d=0$ from origin = $\bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$
The distance of the plane $2x-3y+6z+14=0$ from origin is $\large \frac{14}{\sqrt {4+9+36}}$
$=\large\frac{14}{7}$$=2$
answered Dec 22, 2013

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1 answer