Three Dimensional Geometry

If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane then find value of p. - Wired Faculty

Question

If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane $straight r with rightwards arrow on top. space left parenthesis 3 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 1 space 2 space straight k with hat on top right parenthesis space plus space 13 space equals space 0 comma$ then find value of p.

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 plus space 4 space straight j with hat on top space minus space 12 space straight k with hat on top right parenthesis space plus 13 space equals 0

open parentheses straight x 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 space left parenthesis 3 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 12 space straight k with hat on top right parenthesis space plus 13 space equals space 0

3 straight x plus 4 straight y minus 12 straight z plus 13 space equals space 0

Since the points (1, 1,p) and (–3, 0, 1) are equidistant from this plane.

therefore space space space space fraction numerator open vertical bar 3 plus 4 minus 12 straight p plus 13 close vertical bar over denominator square root of 9 plus 16 plus 144 end root end fraction space equals space fraction numerator open vertical bar negative 9 plus 0 minus 12 plus 13 close vertical bar over denominator square root of 9 plus 16 plus 44 end root end fraction therefore space space space space open vertical bar 20 minus 12 space straight p close vertical bar space equals space open vertical bar negative 8 close vertical bar space space space space space space space space space space rightwards double arrow space space space space open vertical bar 20 space minus space 1 space 2 space straight p close vertical bar space equals space 8 rightwards double arrow space space space space space 20 minus 12 space straight p space equals space plus-or-minus 8 space space space space space space space space space space space space space space rightwards double arrow space space space space space minus 12 space straight p space equals space minus 12 comma space space minus 28 therefore space space space space space space space space space space space space space space straight p space equals space 1 comma space space 7 over 3.