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Three Dimensional Geometry

Question
CBSEENMA12033517

Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the lines:
fraction numerator straight x minus 8 over denominator 3 end fraction space equals space fraction numerator straight y plus 19 over denominator negative 16 end fraction space equals space fraction numerator straight z minus 10 over denominator 7 end fraction space and space fraction numerator straight x minus 15 over denominator 3 end fraction space equals space fraction numerator straight y minus 29 over denominator 8 end fraction space equals space fraction numerator straight z minus 5 over denominator negative 5 end fraction

Solution

Let a. b, c be the direction ratios of the line passing through the point (1, 2, – 4)
∴  equation of line is
fraction numerator straight x minus 1 over denominator straight a end fraction space equals space fraction numerator straight y minus 2 over denominator straight b end fraction space equals space fraction numerator straight z plus 4 over denominator straight c end fraction                           ...(1)
Since this line is perpendicular to the lines
                 fraction numerator straight x minus 8 over denominator 3 end fraction space equals space fraction numerator straight y plus 19 over denominator negative 16 end fraction space equals space fraction numerator straight z minus 10 over denominator 7 end fraction space and space fraction numerator straight x minus 15 over denominator 3 end fraction space equals space fraction numerator straight y minus 29 over denominator 8 end fraction space equals space fraction numerator straight z minus 5 over denominator negative 5 end fraction
therefore space space 3 straight a space minus space 16 straight b space plus space 7 straight c space equals space 0
and    3 straight a plus 8 straight b minus 5 straight c space equals space 0
Solving these equations, we get,
                       fraction numerator straight a over denominator 80 minus 56 end fraction space equals space fraction numerator straight b over denominator 21 plus 15 end fraction equals space fraction numerator straight c over denominator 24 plus 48 end fraction
therefore space space space space space straight a over 24 space equals space straight b over 36 equals straight c over 72
therefore space space space space straight a over 2 space equals space straight b over 3 space equals space straight c over 6
therefore space space space space from space left parenthesis 1 right parenthesis comma space the space equations space of space line space is
fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z plus 4 over denominator 6 end fraction
∴ line passes through the point (1, 2,–4) with position vector straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top is parallel to straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space space 3 space straight j with hat on top space plus space 6 space straight k with hat on top.
∴   equation of line is
                   straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top
or                straight r with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 4 space straight k with hat on top space plus space straight lambda left parenthesis 2 straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis