Verified Solution • 2025–26 Session moderate

QJEEM22Ma0fa7: JEE Main 2022 Mathematics - Class 12 JEEM22Ma0fa7 Solution: If the two lines and are perpendicular, then an angle between…

If the two lines l1:x−23=y+1−2, z=2{l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2 and l2:x−11=2y+3α=z+52{l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2} are perpendicular, then an angle between the lines l2 and l3:1−x3=2y−1−4=z4{l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4} is :

Multiple Choice Options

  • A cos⁡−1(294){\cos ^{ - 1}}\left( {{{29} \over 4}} \right)
  • B sec⁡−1(294){\sec ^{ - 1}}\left( {{{29} \over 4}} \right)
  • C cos⁡−1(229){\cos ^{ - 1}}\left( {{2 \over {29}}} \right)
  • D cos⁡−1(229){\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)

Step-by-Step Verified Solution

∵\because L1 and L2 are perpendicular, so

3×1+(−2)(α2)+0×2=03 \times 1 + ( - 2)\left( {{\alpha \over 2}} \right) + 0 \times 2 = 0

⇒α=3\Rightarrow \alpha = 3

Now angle between l2 and l3,

cos⁡θ=1(−3)+α2(−2)+2(4)1+α24+49+4+16\cos \theta = {{1( - 3) + {\alpha \over 2}( - 2) + 2(4)} \over {\sqrt {1 + {{{\alpha ^2}} \over 4} + } 4\sqrt {9 + 4 + 16} }}

⇒cos⁡θ=2292⇒θ=cos⁡−1(429)=sec⁡−1(294)\Rightarrow \cos \theta = {2 \over {{{29} \over 2}}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{4 \over {29}}} \right) = {\sec ^{ - 1}}\left( {{{29} \over 4}} \right)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring