-->

Three Dimensional Geometry

Question
CBSEENMA12033303

Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
fraction numerator straight x minus 8 over denominator 3 end fraction space 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 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 space space space 3 straight a minus 16 straight b plus 7 straight c space equals space 0
and 3 a + 8 b – 5 c = 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 space equals space fraction numerator straight c over denominator 24 plus 48 end fraction
therefore                     straight a over 24 space equals space straight b over 36 space equals space straight c over 72
therefore space space space space space straight a over 2 space equals space straight b over 3 space equals space straight c over 6
∴   from (1), the equations of line is
fraction numerator straight x minus 1 over denominator 2 end fraction space equals 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 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 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 space left parenthesis 2 space 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

Some More Questions From Three Dimensional Geometry Chapter

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