"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 :"

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Accepted Answer

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$\because$ 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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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 :

Solution

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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 :

Solution

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More Three Dimensional Geometry questions

Drill 25 more like these. Every day. Free.

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 :