Let a and b be two unit vectors such that |(a + b) + 2(a × b)| = 2 . If θ ∈ (0, π) is the angle between ^ a and ^ b , then among the statements: (S1): 2| ^ a × ^ b | = | ^ a - ^ b | (S2) : The projection of ^ a on ( ^ a

Correct answer: C. Both (S1) and (S2) are true. Given, |( ^ a + ^ b ) + 2( ^ a × ^ b )| = 2, θ ∈ (0, π) Squaring both side we get, (( ^ a + ^ b ) + 2( ^ a × ^ b )) · (( ^ a + ^ b ) + 2( ^ a × ^ b )) = 4 | ^ a + ^ b | ^ 2 + 4 |( ^ a × ^ b )| ^ 2 + 0 = 4 Let the angle be θ between ^ a and ^ b ⇒ | ^ a | ^ 2 + | ^ b | ^ 2 + 2| ^ a || ^ b |cosθ + 4(| ^ a || ^ b |sinθ) = 4 ⇒ 2 + 2cosθ + 4 ^ 2 θ = 4 ⇒ 2 + 2cosθ - 4 ^ 2 θ = 0 Let cosθ = t then ⇒ 2 t ^ 2 - t - 1 = 0 ⇒ 2 t ^ 2 - 2t + t - 1 = 0 ⇒ 2t(t - 1) + (t - 1) = 0 (2t + 1)(t - 1) = 0 So, t = - (1)/(2) or t = 1 Then, cosθ = - (1)/(2) or cosθ = 1 which is not possible as 0 ∈ (0, π) So, θ = (2π)/(3) Now, S _ 1 ⇒ 2|a × b| = 2 ((2π)/(3)) = 2 × 3 2 = 3 And | ^ a - ^ b | = 1 + 1 - 2 ((2π)/(3)) = 2 - 2 × (- (1)/(2)) = 3 S _ 1 is correct. S _ 2 projection of ^ a on ( ^ a + ^ b ) ⇒ ^ a . ( ^ a + ^ b ) ∣ ^ a + ^ b ) = 1 + ((2π)/(3)) 2 + 2cos(2π)/(3) ⇒ 1 - (1)/(2) 1 = (1)/(2) Which is true, So both are true.

JEE Main 202224 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 24 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2022 · 24 June · Evening Shift · Mathematics · Vector Algebra

Let a and b be two unit vectors such that |(a + b) + 2(a × b)| = 2...

Let

a and b

be two unit vectors such that \left|\left(a + b\right) + 2\left(a \times b\right)\right| = 2

. If

\theta \in \left(0, \pi\right)

is the angle between

\overset{^}{a} and \overset{^}{b}

, then among the statements:<br />\left(S1\right): 2\left|\overset{^}{a} \times \overset{^}{b}\right| = \left|\overset{^}{a} - \overset{^}{b}\right|<br />\left(S2\right) : The projection of

\overset{^}{a}

on \left(\overset{^}{a} + \overset{^}{b}\right) is \dfrac{1}{2}

Options

AOnly $\left(S1\right)$ is true.
BOnly $\left(S2\right)$ is true.
CBoth $\left(S1\right)$ and $\left(S2\right)$ are true.
DBoth $\left(S1\right)$ and $\left(S2\right)$ are false.

Correct answer

C. Both $\left(S1\right)$ and $\left(S2\right)$ are true.

Step-by-step solution

Given,

$\left|\left(\overset{^}{a} + \overset{^}{b}\right) + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right| = 2, \theta \in \left(0, \pi\right)$

Squaring both side we get,

\left(\left(\overset{^}{a} + \overset{^}{b}\right) + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right) \cdot \left(\left(\overset{^}{a} + \overset{^}{b}\right) + 2\left(\overset{^}{a} \times \overset{^}{b}\right)\right) = 4 $$ {\left|\overset{^}{a} + \overset{^}{b}\right|}^{2} + 4{\left|\left(\overset{^}{a} \times \overset{^}{b}\right)\right|}^{2} + 0 = 4 $Let the angle be$ \theta $between$ \overset{^}{a} $and$ \overset{^}{b} $$ ⇒ {\left|\overset{^}{a}\right|}^{2} + {\left|\overset{^}{b}\right|}^{2} + 2\left|\overset{^}{a}\right|\left|\overset{^}{b}\right|cos\theta + 4\left(\left|\overset{^}{a}\right|\left|\overset{^}{b}\right|sin\theta\right) = 4 $$ ⇒ 2 + 2cos\theta + 4\sin^{2}\theta = 4 $$ ⇒ 2 + 2cos\theta - 4\cos^{2}\theta = 0 $Let$ cos\theta = t $then$ ⇒ 2{t}^{2} - t - 1 = 0 $$ ⇒ 2{t}^{2} - 2t + t - 1 = 0 $$ ⇒ 2t\left(t - 1\right) + \left(t - 1\right) = 0 $\left(2t + 1\right)\left(t - 1\right) = 0$

So, $t = - \dfrac{1}{2}$ or $t = 1$

Then, $cos\theta = - \dfrac{1}{2}$ or $cos\theta = 1$ which is not possible as $0 \in \left(0, \pi\right)$

So, $\theta = \dfrac{2\pi}{3}$

Now,

${S}_{1} ⇒ 2\left|a \times b\right| = 2\sin \left(\dfrac{2\pi}{3}\right) = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3}$

And \left|\overset{^}{a} - \overset{^}{b}\right| = \sqrt{1 + 1 - 2\cos \left(\dfrac{2\pi}{3}\right)} $$ = \sqrt{2 - 2 \times \left(- \dfrac{1}{2}\right)} $$ = \sqrt{3} $$ {S}_{1} $is correct.$ {S}_{2} $projection of$ \overset{^}{a} $on \left(\overset{^}{a} + \overset{^}{b}\right)$

$⇒ \dfrac{\overset{^}{a} . \left(\overset{^}{a} + \overset{^}{b}\right)}{∣ \overset{^}{a} + \overset{^}{b} )} = \dfrac{1 + \cos \left(\dfrac{2\pi}{3}\right)}{\sqrt{2 + 2cos\dfrac{2\pi}{3}}}$

$⇒ \dfrac{1 - \dfrac{1}{2}}{\sqrt{1}} = \dfrac{1}{2}$

Which is true, So both are true.

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