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Three Dimensional Geometry

Question
CBSEENMA12033297
Wired Faculty App

Find the value of p so that the lines
fraction numerator 1 minus straight x over denominator 3 end fraction space equals space fraction numerator 7 straight y minus 14 over denominator 2 space straight p end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction space space and space space fraction numerator 7 minus 7 straight z over denominator 3 space straight p end fraction space equals space fraction numerator straight y minus 5 over denominator 1 end fraction space equals space fraction numerator 6 minus straight z over denominator 5 end fraction

Solution
The equations of two lines are
                           fraction numerator 1 minus straight x over denominator 3 end fraction space equals space fraction numerator 7 straight y minus 14 over denominator 2 space straight p end fraction space equals fraction numerator straight z minus 3 over denominator 2 end fraction space space and space space fraction numerator 7 minus 7 straight x over denominator 3 space straight p end fraction space equals fraction numerator straight y minus 5 over denominator 1 end fraction space equals space fraction numerator 6 minus straight z over denominator 5 end fraction
or                        fraction numerator straight x minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 2 over denominator begin display style fraction numerator 2 straight p over denominator 7 end fraction end style end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction space space and space space fraction numerator straight x minus 1 over denominator negative begin display style fraction numerator 3 straight p over denominator 7 end fraction end style end fraction space equals space fraction numerator straight y minus 5 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
negative 3 comma space space fraction numerator 2 straight p over denominator 7 end fraction comma space 2 space space and space minus fraction numerator 3 straight p over denominator 7 end fraction comma space 1 comma space minus 5
∵   lines are at right angle
therefore         open parentheses negative 3 close parentheses space open parentheses negative fraction numerator 3 space straight p over denominator 7 end fraction close parentheses space plus space open parentheses fraction numerator 2 space straight p over denominator 7 end fraction close parentheses 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 fraction numerator 9 space straight p over denominator 7 end fraction space plus space fraction numerator 2 space straight p over denominator 7 end fraction minus 10 space equals space 0 space space space space space space space space space rightwards double arrow space space space space fraction numerator 11 space straight p over denominator 7 end fraction space equals space 10 therefore space space space space space space space space space straight p space equals space 70 over 11

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.

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