Vector Algebra
Angle Between Two Vectors via Dot Product
nta_pyq_2024_jan
Grade 12
Question:
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:
$\cos^{-1}\left(\dfrac{2}{\sqrt{3}}\right)$
$\cos^{-1}\left(-\dfrac{1}{\sqrt{3}}\right)$
$\cos^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$
$\cos^{-1}\left(\dfrac{2}{3}\right)$
Step-by-Step Solution
Key Concept: Compute $\vec{b}\cdot\vec{c}=\vec{b}\cdot[2(\vec{a}\times\vec{b})-3\vec{b}]=0-3|\vec{b}|^2=-48$. Find $|\vec{c}|^2=4|\vec{a}\times\vec{b}|^2+9|\vec{b}|^2=4(|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2)+9|\vec{b}|^2=4(16-4)+144=192$.
$\vec{b}\cdot\vec{c}=-3|\vec{b}|^2=-48$. $|\vec{c}|^2=4(16-4)+144=192$, $|\vec{c}|=8\sqrt{3}$. $\cos\theta=\dfrac{-48}{4\cdot8\sqrt{3}}=-\dfrac{\sqrt{3}}{2}$. $\theta=\cos^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$.
Correct Answer: 3