Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $|(\hat{a}+\hat{b})+2(\hat{a} \times \hat{b})|=2$. If $\theta \in(0, \pi)$ is the angle between $\hat{a}$ and $\hat{b}$, then among the statements: (S1) : $2|\hat{a} \times \hat{b}|=|\hat{a}-\hat{b}|$ (S2) : The projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{1}{2}$

(C) Both (S1) and (S2) are true

(D) Both (S1) and (S2) are false

Vector Algebra — JEE Main · 2022 · 24 Jun 2022 · Shift 2 Question with Solution

Let

\overset{a}{^}

and

\hat{b}

be two unit vectors such that

∣ (\overset{a}{^} + \hat{b}) + 2 (\overset{a}{^} \times \hat{b}) ∣ = 2

. If

θ \in (0, π)

is the angle between

\overset{a}{^}

and

\hat{b}

, then among the statements:
(S1) :

2∣ \overset{a}{^} \times \hat{b} ∣ = ∣ \overset{a}{^} - \hat{b} ∣

(S2) : The projection of

\overset{a}{^}

(\overset{a}{^} + \hat{b})

\frac{1}{2}

Answer: (C)

Squaring the given condition:

∣ \overset{a}{^} + \hat{b} ∣^{2} + 4∣ \overset{a}{^} \times \hat{b} ∣^{2} = 4

. Since

∣ \overset{a}{^} + \hat{b} ∣^{2} = 2 + 2 cos θ

and

∣ \overset{a}{^} \times \hat{b} ∣^{2} = sin^{2} θ

, we get

2 + 2 cos θ + 4 sin^{2} θ = 4

. Using

sin^{2} θ = 1 - cos^{2} θ

2 cos^{2} θ - cos θ - 1 = 0 \Rightarrow (2 cos θ + 1) (cos θ - 1) = 0

. Since

θ \in (0, π)

cos θ = - \frac{1}{2}

, so

θ = \frac{2 π}{3}

. For (S1):

2∣ \overset{a}{^} \times \hat{b} ∣ = 2 sin \frac{2 π}{3} = 3

and

∣ \overset{a}{^} - \hat{b} ∣ = 2 - 2 cos \frac{2 π}{3} = 3

✓. For (S2): projection =

\frac{a ^ \cdot ( a ^ + b ^ )}{∣ a ^ + b ^ ∣} = \frac{1 + c o s \frac{2 π}{3}}{2 + 2 c o s \frac{2 π}{3}} = \frac{1/2}{1} = \frac{1}{2}

✓. Both true.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option