JEE Main 202226 Jun, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002If the two lines l1:x−23=y+1−2, z=2{l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2l1:3x−2=−2y+1,z=2 and l2:x−11=2y+3α=z+52{l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}l2:1x−1=α2y+3=2z+5 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}l3:31−x=−42y−1=4z is :AOption A. cos−1(294){\cos ^{ - 1}}\left( {{{29} \over 4}} \right)cos−1(429)BOption B. sec−1(294){\sec ^{ - 1}}\left( {{{29} \over 4}} \right)sec−1(429)COption C. cos−1(229){\cos ^{ - 1}}\left( {{2 \over {29}}} \right)cos−1(292)DOption D. cos−1(229){\cos ^{ - 1}}\left( {{2 \over {\sqrt {29} }}} \right)cos−1(292)CheckSkip · show solution
