Question

Find the length of the perpendicular drawn from the origin to the plane $2x-3y+6z+21=0$.

Answer 1

Given a plane $2x-3y+6z+21=0$

We know the formula, the distance of a point $P(x_1,y_1,z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $\frac{A{x}_1+B{y}_1+C{z}_1+D}{\sqrt{A^2+B^2+C^2}}$

We need to calculate the distance from the origin P(0, 0, 0) to the place.

Distance $=\frac{2\left(0\right)-3(0+6(0)+21)}{\sqrt{2^2+3^2+6^2}}$

$=\frac{21}{\sqrt{49}}=\frac{21}{7}=3$

Therefore, the length of the perpendicular drawn from the origin to the plane $2x–3y+6z+21=0$ is $3.$