-->

Three Dimensional Geometry

Question
CBSEENMA12033295

Find the angle between the following pair of lines:
fraction numerator straight x minus 5 over denominator 1 end fraction equals space fraction numerator 2 straight y plus 6 over denominator negative 2 end fraction space equals space fraction numerator straight z minus 3 over denominator 1 end fraction space space and space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction
 


Solution
The equations of two lines are
                    fraction numerator straight x minus 5 over denominator 1 end fraction space equals space fraction numerator 2 straight y plus 6 over denominator negative 2 end fraction space equals space fraction numerator straight z minus 3 over denominator 1 end fraction space and space space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction
or               fraction numerator straight x minus 5 over denominator 1 end fraction space equals space fraction numerator straight y plus 3 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 3 over denominator 1 end fraction space space and space space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 6 over denominator 5 end fraction
Direction ratios of two lines are 1, –1, 1 and 3, 4, 5.
Let θ be angle between the lines.
therefore space space space space cos space straight theta space equals space fraction numerator left parenthesis 1 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 4 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis 5 right parenthesis over denominator square root of 1 plus 1 plus 1 end root space square root of 9 plus 16 plus 25 end root end fraction space equals space fraction numerator 3 minus 4 plus 5 over denominator square root of 3 space square root of 50 end fraction
                    equals space fraction numerator 4 over denominator square root of 3 space cross times space 5 square root of 2 end fraction space equals space fraction numerator 4 over denominator 5 square root of 6 end fraction
therefore space space space space straight theta space equals space cos to the power of negative 1 end exponent space open parentheses fraction numerator 4 over denominator 5 square root of 6 end fraction close parentheses

Some More Questions From Three Dimensional Geometry Chapter