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Three Dimensional Geometry

Question
CBSEENMA12033468

Find the angle between the line straight x over 4 space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 1 over denominator 6 end fraction and the plane x + 2y + 2 + 3 = 0

Solution
The equation of line is
              straight x over 4 space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 1 over denominator 6 end fraction
Its direction ratios are 4, 3, 6
The equation of plane is
x + 2y + z + 3 = 0
∴  direction ratios of the normal to the plane are 1,2, 1.
Let θ be the angle between the line and plane
therefore space space space space sin space straight theta space equals space fraction numerator left parenthesis 4 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis 3 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis 6 right parenthesis thin space left parenthesis 1 right parenthesis over denominator square root of 16 plus 9 plus 36 end root space square root of 1 plus 4 plus 1 end root end fraction space equals space fraction numerator 4 plus 6 plus 6 over denominator square root of 61 space square root of 6 end fraction space equals space fraction numerator 16 over denominator square root of 366 end fraction
therefore space space space space space space space space straight theta space equals space sin to the power of negative 1 end exponent space open parentheses fraction numerator 16 over denominator square root of 366 end fraction close parentheses.