14 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

c. (2, 3, – 5)

x + 2y – 2z = 9

d. (– 6, 0, 0)

2x – 3y + 6z – 2 = 0

 

Answers (1)

We know that the distance between a point P (x_{1},y_{1},z_{1}) and a plane Ax+By+Cz =D is given by,

d =\left | \frac{Ax_{1}+By_{1}+Cz_{1}-D}{\sqrt{A^2+B^2+C^2}} \right |                                   .......................(1)

So, calculating for each case;

(a) Point (0,0,0) and Plane 3x-4y+12z = 3

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

(b) Point (3,-2,1) and Plane 2x-y+2z +3= 0

Therefore, d =\left | \frac{2(3)-(-2)+2(1)+3}{\sqrt{2^2+(-1)^2+2^2}} \right | = \frac{13}{3}

(c) Point (2,3,-5) and Plane x+2y-2z =9

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

(d) Point (-6,0,0) and Plane 2x-3y+6z -2= 0

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

 

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