If the two lines ${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$ and ${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$ are perpendicular, then an angle between the lines l2 and ${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$ is :

Question

If the two lines ${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$ and ${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$ are perpendicular, then an angle between the lines l₂ and ${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$ is :

${\cos ^{ - 1}}\left( {{{29} \over 4}} \right)$

${\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$

${\cos ^{ - 1}}\left( {{2 \over {29}}} \right)$

${\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)$

Solution

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