# In the following cases, find the distance of each of the given points from the corresponding given plane.

Question:

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0)

(b) (3, −2, 1)

(c) (2, 3, −5)

(d) (−6, 0, 0)

Solution:

It is known that the distance between a point, p(x1y1z1), and a plane, Ax + By + Cz = D, is given by,

$d=\left|\frac{A x_{1}+B y_{1}+C z_{1}-D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|$            ...(1)

(a) The given point is (0, 0, 0) and the plane is

$\therefore d=\left|\frac{3 \times 0-4 \times 0+12 \times 0-3}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}}\right|=\frac{3}{\sqrt{169}}=\frac{3}{13}$

(b) The given point is (3, − 2, 1) and the plane is

$\therefore d=\left|\frac{2 \times 3-(-2)+2 \times 1+3}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}\right|=\left|\frac{13}{3}\right|=\frac{13}{3}$

(c) The given point is (2, 3, −5) and the plane is

$\therefore d=\left|\frac{2+2 \times 3-2(-5)-9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}\right|=\frac{9}{3}=3$

(d) The given point is (−6, 0, 0) and the plane is

$d=\left|\frac{2(-6)-3 \times 0+6 \times 0-2}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}}\right|=\left|\frac{-14}{\sqrt{49}}\right|=\frac{14}{7}=2$