Question

Find the distance of the point (2, 3, 5) from the xy - plane.

Answer 1

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