Three Dimensional Geometry

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Question

Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, –1, 2), B(5, 2, 4) and C(– 1, – 1, 6).

Solution

The equation of plane through the points (3, –1, 2), (5, 2, 4), (– 1, – 1, 6) is

open vertical bar table row cell straight x minus 3 end cell cell straight y plus 1 end cell cell straight z minus 2 end cell row cell 5 minus 3 end cell cell 2 plus 1 end cell cell 4 minus 2 end cell row cell negative 1 minus 3 end cell cell negative 1 plus 1 end cell cell 6 minus 2 end cell end table close vertical bar space equals 0 space space space space space space space space space space space open square brackets open vertical bar table row cell straight x minus straight x subscript 1 end cell cell straight y minus straight y subscript 1 end cell cell straight z minus straight z subscript 1 end cell row cell straight x subscript 2 minus straight x subscript 1 end cell cell straight y subscript 2 minus straight y subscript 1 end cell cell straight z subscript 2 minus straight z subscript 1 end cell row cell straight x subscript 3 minus straight x subscript 1 end cell cell straight y subscript 3 minus straight y subscript 1 end cell cell straight z subscript 3 minus straight z subscript 1 end cell end table close vertical bar space equals space 0 close square brackets

open vertical bar table row cell straight x minus 3 end cell cell straight y plus 1 end cell cell straight z minus 2 end cell row 2 3 2 row cell negative 4 end cell 0 4 end table close vertical bar space equals space 0

or space space space space left parenthesis straight x minus 3 right parenthesis space open vertical bar table row 3 2 row 0 4 end table close vertical bar space minus space left parenthesis straight y plus 1 right parenthesis space open vertical bar table row 2 2 row cell negative 4 end cell 4 end table close vertical bar space plus space left parenthesis straight z minus 2 right parenthesis space open vertical bar table row 2 3 row cell negative 4 end cell 0 end table close vertical bar space equals space 0

or space space space space space space left parenthesis straight x minus 3 right parenthesis thin space left parenthesis 12 minus 0 right parenthesis space minus space left parenthesis straight y plus 1 right parenthesis space left parenthesis 8 plus 8 right parenthesis space plus space left parenthesis straight z minus 2 right parenthesis space left parenthesis 0 plus 12 right parenthesis space equals space 0 or space space space space space space 12 space left parenthesis straight x minus 3 right parenthesis space minus space 16 space left parenthesis straight y plus 1 right parenthesis space plus space 12 space left parenthesis straight z minus 2 right parenthesis space equals space 0 or space space space space space space 3 left parenthesis straight x minus 3 right parenthesis space space minus space 4 space left parenthesis straight y plus 1 right parenthesis space plus space 3 space left parenthesis straight z minus 2 right parenthesis space equals space 0 or space space space space space space 3 straight x minus 9 minus 4 straight y minus 4 plus 3 straight z minus 6 space equals space 0 or space space space space space 3 straight x minus 4 straight y plus 3 straight z minus 19 space equals space 0

Let p be the perpendicular distance of P(6, 5, 9) from this plane.

therefore space space space space space space straight p space equals space fraction numerator 3 left parenthesis 6 right parenthesis minus 4 left parenthesis 5 right parenthesis space plus space 3 left parenthesis 9 right parenthesis space minus space 19 over denominator square root of 9 plus 16 plus 9 end root end fraction space equals fraction numerator 18 minus 20 plus 27 minus 19 over denominator square root of 34 end fraction equals space fraction numerator 6 over denominator square root of 34 end fraction