Three Dimensional Geometry

Find the distance of the point (2, 3, 4) from the plane - Wired Faculty

Question

Find the distance of the point (2, 3, 4) from the plane
$straight r with rightwards arrow on top. space open parentheses 3 space straight i with hat on top space minus space 6 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space equals space minus 11.$

Solution

The equation of the plane is

straight r with rightwards arrow on top. space left parenthesis 3 space straight i with hat on top space minus space 6 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis space equals space minus 11

or space space left parenthesis straight x straight i with hat on top space plus space straight y straight j with hat on top space space plus space straight z straight k with hat on top right parenthesis. space space left parenthesis 3 straight i with hat on top space minus space 6 straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis equals space minus 11 or space space 3 straight x minus 6 straight y plus 2 straight z space equals space minus 11 or space space 3 straight x minus 6 straight y plus 2 straight z plus 11 space equals space 0

Let p be the length of perpendicular from (2, 3,4) to the plane.

therefore space space space space straight p space space equals space fraction numerator 3 left parenthesis 2 right parenthesis space minus space 6 left parenthesis 3 right parenthesis space plus space 2 left parenthesis 4 right parenthesis space plus space 11 over denominator square root of left parenthesis 3 right parenthesis squared space plus space left parenthesis negative 6 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root end fraction space space space space space space space space space space space space equals space fraction numerator 6 minus 18 plus 8 plus 11 over denominator square root of 9 plus 36 plus 4 end root end fraction space equals space fraction numerator 7 over denominator square root of 49 end fraction space equals space 7 over 7 space equals space 1