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Three Dimensional Geometry

Question
CBSEENMA12033299

If the lines fraction numerator straight x minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 2 straight k end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction space and space fraction numerator straight x minus 1 over denominator 3 straight k end fraction space equals space fraction numerator straight y minus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 5 end fraction are prependicular, then find the value of k.

Solution
The given lines are
                 fraction numerator straight x minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 2 space straight k end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction
and      fraction numerator straight x minus 1 over denominator 3 space straight k end fraction space equals space fraction numerator straight y minus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 5 end fraction
Direction ratios of two lines are – 3, 2k, 2 and 3 k, 1, –5
Since the lines are perpendicular
therefore space space space left parenthesis negative 3 right parenthesis thin space left parenthesis 3 straight k right parenthesis space plus space left parenthesis 2 straight k right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 2 right parenthesis thin space left parenthesis negative 5 right parenthesis space equals space 0
therefore space space space space minus 9 straight k space plus space 2 straight k space minus space 10 space equals space 0 space space space space space space space space space space space space rightwards double arrow space minus 7 space straight k space equals space 10
therefore space space space space space straight k space equals space minus 10 over 7