Verified Solution • 2025–26 Session moderate

QJEEM21Ma74ca: JEE Main 2021 Physics - Class 11 JEEM21Ma74ca Solution: If vectors and are collinear, then a possible unit vector…

If vectors a1→=xi^−j^+k^\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k and a2→=i^+yj^+zk^\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k are collinear, then a possible unit vector parallel to the vector xi^+yj^+zk^x\widehat i + y\widehat j + z\widehat k is :

Multiple Choice Options

  • A 13(i^−j^+k^){1 \over {\sqrt 3 }}\left( {\widehat i - \widehat j + \widehat k} \right)
  • B 12(−j^+k^){1 \over {\sqrt 2 }}\left( { - \widehat j + \widehat k} \right)
  • C 12(i^−j^){1 \over {\sqrt 2 }}\left( {\widehat i - \widehat j} \right)
  • D 13(i^+j^−k^){1 \over {\sqrt 3 }}\left( {\widehat i + \widehat j - \widehat k} \right)

Step-by-Step Verified Solution

a2→=λa1→\overrightarrow {{a_2}} = \lambda \overrightarrow {{a_1}}

i^+yj^+zk^=λ(xi^−j^+k^)\widehat i + y\widehat j + z\widehat k = \lambda (x\widehat i - \widehat j + \widehat k)

1=λx,y=−λ,z=λ1 = \lambda x,y = - \lambda ,z = \lambda

xi^+yj^+zk^=1λi^−λj^+λk^x\widehat i + y\widehat j + z\widehat k = {1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k

Unit vector =1λi^−λj^+λk^1λ2+λ2+λ2= {{{1 \over \lambda }\widehat i - \lambda \widehat j + \lambda \widehat k} \over {\sqrt {{1 \over {{\lambda ^2}}} + {\lambda ^2} + {\lambda ^2}} }}

=i^−λ2j^+λ2k^1+2λ4= {{\widehat i - {\lambda ^2}\widehat j + {\lambda ^2}\widehat k} \over {\sqrt {1 + 2{\lambda ^4}} }}

Let λ2=1{\lambda ^2} = 1 , possible unit vector =i^−j^+k^3= {{\widehat i - \widehat j + \widehat k} \over {\sqrt 3 }}

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