Let $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$ and $\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$ be two vectors such that $|\vec{a}|=1$; $\vec{a} \cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to :

(C) $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

(A) $\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

(B) $\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

(D) $\cos ^{-1}\left(\frac{2}{3}\right)$

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Vector Algebra

Let

a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}

and

b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

be two vectors such that

∣ a ∣ = 1

;

a \cdot b = 2

and

∣ b ∣ = 4

. If

c = 2 (a \times b) - 3 b

, then the angle between

b

and

c

is equal to :

Answer: (C)

b \cdot c = 2 b \cdot (a \times b) - 3∣ b ∣^{2} = 0 - 3 (16) = - 48

∣ c ∣^{2} = 4∣ a \times b ∣^{2} + 9∣ b ∣^{2} = 4 (∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}) + 9 (16) = 4 (16 - 4) + 144 = 192

, so

∣ c ∣ = 83

cos θ = \frac{b \cdot c}{∣ b ∣∣ c ∣} = \frac{- 48}{4 ( 8 3 )} = - \frac{3}{2}

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option