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Three Dimensional Geometry

Question
CBSEENMA12033459
Wired Faculty App

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 A(3, –1, 2), B(5, 2, 4), C(–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 space 0
or       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 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 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 thin space left parenthesis 0 plus 12 right parenthesis space equals space 0 or space space 12 left parenthesis straight x minus 3 right parenthesis space minus space 16 space left parenthesis straight y plus 1 right parenthesis space plus 12 space left parenthesis straight z minus 2 right parenthesis space equals space 0 or space space 3 space left parenthesis straight x minus 3 right parenthesis 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 3 straight x minus 9 space minus space 4 straight y space minus 4 space plus space 3 straight z space minus space 6 space equals space space 0 or space space 3 straight x minus 4 straight y plus 3 straight z minus 19 space equals space 0
Let p be distance between P(6, 5, 9) and plane.
therefore space space space space straight p space equals space fraction numerator 3 space left parenthesis 6 right parenthesis space minus space 4 space left parenthesis 5 right parenthesis space plus space 3 space left parenthesis 9 right parenthesis space minus space 19 over denominator square root of left parenthesis 3 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus left parenthesis 3 right parenthesis squared end root end fraction space equals space fraction numerator 18 minus 20 plus 27 minus 19 over denominator square root of 9 plus 16 plus 9 end root end fraction
            equals space fraction numerator 6 over denominator square root of 34 end fraction space equals space fraction numerator 6 square root of 34 over denominator 34 end fraction space equals space fraction numerator 3 square root of 34 over denominator 17 end fraction

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin2α + sin2 β + sin2γ = 2.

Show that the points A(2, 3, – 4), B(1, –2, 3) and C (3, 8, –11) are collinear.

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