Vector Algebra Question 9
Question 9 - 2024 (30 Jan Shift 1)
Let $\vec{a}=a_{i} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$ and $\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k} \quad$ be two vectors such that $|\vec{a}|=1 ; \quad \vec{a} \cdot \vec{b}=2$ 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:
(1) $\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
(2) $\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
(3) $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
(4) $\cos ^{-1}\left(\frac{2}{3}\right)$
Show Answer
Answer (3)
Solution
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
$\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-48$
$\vec{c} \cdot \vec{c}=4|\vec{a} \times \vec{b}|^{2}+9|\vec{b}|^{2}$
$|\overrightarrow{\mathrm{c}}|^{2}=4\left[|\mathrm{a}|^{2}|\mathrm{~b}|^{2}-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^{2}\right]+9|\overrightarrow{\mathrm{b}}|^{2}$
$|\overrightarrow{\mathrm{c}}|^{2}=4\left[(1)(4)^{2}-(4)\right]+9(16)$
$|\overrightarrow{\mathrm{c}}|^{2}=4[12]+144$
$|\overrightarrow{\mathrm{c}}|^{2}=48+144$
$|\overrightarrow{\mathrm{c}}|^{2}=192$
$\therefore \cos \theta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}$
$\therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4}$
$\therefore \cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4}$
$\therefore \cos \theta=\frac{-3}{2 \sqrt{3}}$
$\therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
