In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane (a) (0, 0, 0) 3x – 4y + 12 z = 3 (b) (3, -2, 1) 2x

Home » In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane (a) (0, 0, 0) 3x – 4y + 12 z = 3 (b) (3, -2, 1) 2x – y + 2z + 3 = 0

In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3
(b) (3, -2, 1) 2x – y + 2z + 3 = 0

In the following cases, find the distance of each of the given points from the corresponding given plane. Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3
(b) (3, -2, 1) 2x – y + 2z + 3 = 0

Solution:

(a) The distance of the point $(0,0,0)$ from the plane $3 x-4 y+12=3 \Rightarrow$ $3 x-4 y+12 z-3=0$ is
$\begin{array}{l} \frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} \\ =\frac{|3(0)-4(0)+12(0)-3|}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}} \\ =\frac{|3|}{\sqrt{9+16+144}}=\frac{3}{\sqrt{169}}=\frac{3}{13} \end{array}$

(b) The length of perpendicular from the point $(3,-2,1)$ on the plane $2 x-y+2 z+3=0$ is
$\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$=\frac{|2(3)-(-2)+2(1)+3|}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}$
$=\frac{|13|}{\sqrt{4+1+4}}=\frac{13}{\sqrt{9}}=\frac{13}{3}$