JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 10)
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 :
$${\cos ^{ - 1}}\left( {{{29} \over 4}} \right)$$
$${\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$$
$${\cos ^{ - 1}}\left( {{2 \over {29}}} \right)$$
$${\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)$$
Explanation
$$\because$$ L1 and L2 are perpendicular, so
$$3 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0$$
$$ \Rightarrow \alpha = 3$$
Now angle between l2 and l3,
$$\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over {\sqrt {1 + {{{\alpha ^2}} \over 4} + } 4\sqrt {9 + 4 + 16} }}$$
$$ \Rightarrow \cos \theta = {2 \over {{{29} \over 2}}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right) = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)$$
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