Jee Main Jan 30 Shift 1 Mathematics - Question13

If $|\vec{a}|=1,|\vec{b}|=4$

$\vec{a} \cdot \vec{b}=2$ and $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$

Then the angle between $\vec{b}$ and $\vec{c}$ is

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

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

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

(4) $\theta=\cos ^{-1}\left(\frac{-1}{2}\right)$

Show Answer

Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$

$$ \vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b} $$

Dot product with $\vec{a}$ on both sides

$\vec{c} \cdot \vec{a}=-6$

Dot product with $\vec{b}$ on both sides

$\vec{b} \cdot \vec{c}=-48$

$\vec{c} \cdot \vec{c}=4|\vec{a} \times \vec{b}|^{2}+9|\vec{b}|^{2}$

$|\vec{c}|^{2}=4\left[|\vec{a}|^{2} \cdot|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}\right]+9|\vec{b}|^{2}$

$|\vec{c}|^{2}=4\left[(1)(4)^{2}-(4)\right]+9(16)$ $|\vec{c}|^{2}=4[12]+144$

$|\vec{c}|^{2}=48+144$

$|\vec{c}|^{2}=192$

$\therefore \quad \cos \theta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}$

$\cos \theta=\frac{-48}{\sqrt{192} \cdot 4}$

$\cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4}$

$\cos \theta=\frac{-3}{2 \sqrt{3}}$

$\cos \theta=\frac{-\sqrt{3}}{2} \quad \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$