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