Question

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.

Answer 1

$$\begin{gathered} \text{We know that the distance of the point}\left(x_1,y_1,z_1\right)\text{from the plane}ax+by+cz+d=0\text{is given by} \\ \frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}} \\ \mathbf{\text{Distance of the point (1, 1, 1) from the plane 3x + 4y - 12z + 13 = 0}} \\ \text{The required distance} \\ =\frac{\left|3\left(1\right)+4\left(1\right)-12\left(1\right)+13\right|}{\sqrt{3^2+4^2+{\left(-12\right)}^2}} \\ =\frac{\left|3+4-12+13\right|}{\sqrt{9+16+144}} \\ =\frac{8}{13}\text{units ... (1)} \\ \mathbf{\text{Distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0}} \\ \text{The required distance} \\ =\frac{\left|3\left(-3\right)+4\left(0\right)-12\left(1\right)+13\right|}{\sqrt{3^2+4^2+{\left(-12\right)}^2}} \\ =\frac{\left|-9+0-12+13\right|}{\sqrt{9+16+144}} \\ =\frac{8}{13}\text{units .... (2)} \\ \text{From (1) and (2), we can say that the given points are equidistant from the given plane.} \end{gathered}$$