JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 13)
Let $$\widehat a$$ and $$\widehat b$$ be two unit vectors such that $$|(\widehat a + \widehat b) + 2(\widehat a \times \widehat b)| = 2$$. If $$\theta$$ $$\in$$ (0, $$\pi$$) is the angle between $$\widehat a$$ and $$\widehat b$$, then among the statements :
(S1) : $$2|\widehat a \times \widehat b| = |\widehat a - \widehat b|$$
(S2) : The projection of $$\widehat a$$ on ($$\widehat a$$ + $$\widehat b$$) is $${1 \over 2}$$
Explanation
$$\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right| = 2,\,\theta \in (0,\,\pi )$$
$$ \Rightarrow {\left| {\widehat a + \widehat b + 2(\widehat a \times \widehat b)} \right|^2} = 4$$
$$ \Rightarrow {\left| {\widehat a} \right|^2} + {\left| {\widehat b} \right|^2} + 4{\left| {\widehat a \times \widehat b} \right|^2} + 2\widehat a\,.\,\widehat b = 4$$
$$\therefore$$ $$\cos \theta = \cos 2\theta $$
$$\therefore$$ $$\theta = {{2\pi } \over 3}$$
where $$\theta$$ is angle between $$\widehat a$$ and $$\widehat b$$.
$$\therefore$$ $$2\left| {\widehat a \times \widehat b} \right| = \sqrt 3 = \left| {\widehat a - \widehat b} \right|$$
(S1) is correct.
And projection of $$\widehat a$$ on $$(\widehat a + \widehat b) = \left| {{{\widehat a\,.\,(\widehat a + \widehat b)} \over {\left| {\widehat a + \widehat b} \right|}}} \right| = {1 \over 2}$$
(S2) is correct.
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