Question

Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.

Answer 1

$$\begin{gathered} \text{We know that the distance of the point}\left(x_1,y_1,z_1\right)\text{from the plane}ax+by+cz+d=0\text{is given by} \\ \frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the required distance} \\ =\frac{\left|\left(2\right)+2\left(3\right)-2\left(-5\right)-9\right|}{\sqrt{1^2+2^2+{\left(-2\right)}^2}} \\ =\frac{\left|2+6+10-9\right|}{\sqrt{1+4+4}} \\ =\frac{9}{3} \\ =3\text{units} \end{gathered}$$