Three Dimensional Geometry

Find the distance of a point (2, 5, -3) from the plane - Wired Faculty

Question

Find the distance of a point (2, 5, -3) from the plane $straight r with rightwards arrow on top. space open parentheses 6 straight i with hat on top space minus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space equals space 4.$

Solution

The equation of plane is

straight r with rightwards arrow on top. space open parentheses 6 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space equals space 4

open parentheses straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top close parentheses. space open parentheses 6 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space equals space 4

6 straight x minus 3 straight y plus 2 straight z minus 4 space equals space 0

Let p be the length of perpendicular from (2, 5, - 3) to the plane.

therefore space space space space space space space space space straight p space equals space fraction numerator open vertical bar 6 space left parenthesis 2 right parenthesis space minus space 3 space left parenthesis 5 right parenthesis space plus space 2 space left parenthesis negative 3 right parenthesis space minus space 4 close vertical bar over denominator square root of left parenthesis 6 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root end fraction space equals space fraction numerator open vertical bar 12 minus 15 minus 6 minus 4 close vertical bar over denominator square root of 36 plus 9 plus 4 end root end fraction space equals fraction numerator open vertical bar negative 13 close vertical bar over denominator square root of 49 end fraction space equals space 13 over 7