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) 3x-4y+12z=3
(b) (3,-2,1) 2x-y+2z+3=0
(c) (2,3,-5) x+2y-2z=9
(d) (-6,0,0) 2x-3y+6z-2=0

Answer 1

It is known that the distance between a point P(x1,y1,z1)and a plane Ax-+By+Cz=D, is given by,
d=$\begin{vmatrix}\frac{AX_1+By_1+Cz_1-D}{\sqrt{A^2+B^2+C^2}} \end{vmatrix}$|...(1)

(a) The given point is (0,0,0) and the plane is 3x-4y+12z=3

∴d=$\begin{vmatrix}\frac{3*0-4*0+12*0-3}{\sqrt{(3)^2+(-4)^2+(12)^2}} \end{vmatrix}$

=$\frac{3}{\sqrt{169}}$=$\frac{3}{13}$

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(b)The given point is (3,-2,1) and the plane is 2x-y+2z+3=0

∴d=$\begin{vmatrix}\frac{2*3-(-2)+2*1+3}{\sqrt{(2)^2+(-1)^2+(2)^2}} \end{vmatrix}$

=|$\frac{13}{3}$| =$\frac{13}{3}$

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(c)The given point is (2,3,-5) and the plane is x+2y-2z=9

∴d=$\begin{vmatrix}\frac{2+2*3-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}} \end{vmatrix}$

=$\frac{9}{3}$
=3

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(d)The given point is (-6,0,0) and the plane is 2x-3y+6z-2=0

∴d=$\begin{vmatrix}\frac{2(-6)-3*0+6*0-2}{\sqrt{(2)^2+(-3)^2+(6)^2}} \end{vmatrix}$

=$\begin{vmatrix}\frac{-14}{\sqrt{49}} \end{vmatrix}$

=$\frac{14}{7}$
=2