Verified Solution • 2025–26 Session moderate

QJEEM22Ma71f8: JEE Main 2022 Physics - Class 11 JEEM22Ma71f8 Solution: Let and be two unit vectors such that . If (0, ) is the angle…

Let a^\widehat a and b^\widehat b be two unit vectors such that ∣(a^+b^)+2(a^×b^)∣=2|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2 . If θ\theta ∈\in (0, π\pi ) is the angle between a^\widehat a and b^\widehat b , then among the statements :

(S1) : 2∣a^×b^∣=∣a^−b^∣2|\widehat a \times \widehat b| = |\widehat a - \widehat b|

(S2) : The projection of a^\widehat a on ( a^\widehat a + b^\widehat b ) is 12{1 \over 2}

Multiple Choice Options

  • A Only (S1) is true.
  • B Only (S2) is true.
  • C Both (S1) and (S2) are true.
  • D Both (S1) and (S2) are false.

Step-by-Step Verified Solution

∣a^+b^+2(a^×b^)∣=2, θ∈(0, π)\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )

⇒∣a^+b^+2(a^×b^)∣2=4\Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4

⇒∣a^∣2+∣b^∣2+4∣a^×b^∣2+2a^ . b^=4\Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\widehat a \times \widehat b} \right|^2} + 2\widehat a\,.\,\widehat b = 4

∴\therefore cos⁡θ=cos⁡2θ\cos \theta = \cos 2\theta

∴\therefore θ=2π3\theta = {{2\pi } \over 3}

where θ\theta is angle between a^\widehat a and b^\widehat b .

∴\therefore 2∣a^×b^∣=3=∣a^−b^∣2\left| {\widehat a \times \widehat b} \right| = \sqrt 3 = \left| {\widehat a - \widehat b} \right|

(S1) is correct.

And projection of a^\widehat a on (a^+b^)=∣a^ . (a^+b^)∣a^+b^∣∣=12(\widehat a + \widehat b) = \left| {{{\widehat a\,.\,(\widehat a + \widehat b)} \over {\left| {\widehat a + \widehat b} \right|}}} \right| = {1 \over 2}

(S2) is correct.

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