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Three Dimensional Geometry

Question
CBSEENMA12033420

Find the vector equation of the plane passing through the intersection of the planes straight r with rightwards arrow on top. space open parentheses 2 straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space equals 7 comma space space space space space straight r with rightwards arrow on top space. space open parentheses 2 space straight i with hat on top space plus space 5 space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space equals space 9 and the point (2, 1, 3).

Solution
The equations of given planes are
               straight r with rightwards arrow on top. space left parenthesis 2 space straight i with hat on top space plus space straight j with hat on top space plus space space 3 space straight k with hat on top right parenthesis space equals space 7
i.e.   left parenthesis 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 right parenthesis. space space left parenthesis 2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 7
i.e.    2 straight x plus straight y plus 3 straight z minus 7 space equals space 0                                          ...(1)
and straight r with rightwards arrow on top. space left parenthesis 2 straight i with hat on top space plus space 5 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 9
i.e. left parenthesis 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 right parenthesis. space space space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top close parentheses space equals space 9
i.e. 2 straight x plus 5 straight y plus 3 straight z minus 9 space equals space 0                                          ...(2)
Any plane through the intersection of (1) and (2) is
(2x + y + 3z – 7) + k (2x + 5y + 3z – 9) = 0    ...(3)
∴    it passes through (2, 1, 3)
therefore space space space left parenthesis 4 plus 1 plus 9 minus 7 right parenthesis space plus space straight k left parenthesis 4 plus 5 plus 9 minus 9 right parenthesis equals space 0
therefore space space space space 7 plus 9 space straight k space equals space 0 space space space space space space space space space rightwards double arrow space space space space space straight k space equals space minus 7 over 9
Putting straight k space equals space minus 7 over 9 in (3), we get,
          left parenthesis 2 straight x plus straight y plus 3 straight z minus 7 right parenthesis space minus 7 over 9 left parenthesis 2 straight x plus 5 straight y plus 3 straight z minus 9 right parenthesis space equals space 0
           or space space space 9 space left parenthesis 2 straight x plus straight y plus 3 straight z minus 7 right parenthesis space minus space 7 space left parenthesis 2 straight x plus 5 straight y plus 3 straight z minus 9 right parenthesis space equals space 0
or space space 18 straight x plus 9 straight y plus 27 straight z minus 63 minus 14 straight x minus 35 straight y minus 21 straight z plus 63 space equals 0
or space space 4 straight x minus 26 straight y plus 6 straight z space equals space 0 space space space space space space space or space space space 2 straight x minus 13 straight y plus 3 straight z space equals space 0
or space space space space space left parenthesis straight x straight i with hat on top space plus space straight y straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space space left parenthesis 2 space straight i with hat on top space minus space 13 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 0
or space space space straight r with rightwards arrow on top. space left parenthesis 2 straight i with hat on top space minus space 13 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 0
which is vector equation of required plane. 

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.