Verified Solution • 2025–26 Session moderate

QJEEM21Maa01a: JEE Main 2021 Mathematics - Class 12 JEEM21Maa01a Solution: Let be the angle between the lines whose direction cosines…

Let α\alpha be the angle between the lines whose direction cosines satisfy the equations l + m −- n = 0 and l2 + m2 −- n2 = 0. Then the value of sin4 α\alpha + cos4 α\alpha is :

Multiple Choice Options

  • A 38{{3 \over 8}}
  • B 34{{3 \over 4}}
  • C 12{{1 \over 2}}
  • D 58{{5 \over 8}}

Step-by-Step Verified Solution

l2+m2+n2=1{l^2} + {m^2} + {n^2} = 1

∴\therefore 2n2=12{n^2} = 1 ( ∵\because l2 + m2 −- n2 = 0)

⇒n=±12\Rightarrow n = \pm {1 \over {\sqrt 2 }}

∴\therefore l2+m2=12{l^2} + {m^2} = {1 \over 2} & l+m=12l + m = {1 \over {\sqrt 2 }}

⇒12−2lm=12\Rightarrow {1 \over 2} - 2lm = {1 \over 2}

⇒lm=0\Rightarrow lm = 0 or m=0m = 0

∴\therefore l=0,m=12l = 0,m = {1 \over {\sqrt 2 }} or l=12l = {1 \over {\sqrt 2 }}

< 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} > or < {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} >

∴\therefore cos⁡α=0+0+12=12\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}

∴\therefore sin⁡4α+cos⁡4α=1−12sin⁡2(2α)=1−12.34=58{\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}

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