Vector Algebra
Angle Between Vectors Defined via Cross Product
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1$, $|\vec{b}|=4$ and $\vec{a}\cdot\vec{b}=2$. If $\vec{c}=(2\vec{a}\times\vec{b})-3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then $192\sin^2\alpha$ is equal to _____.
Step-by-Step Solution
Key Concept: Find $\vec{b}\cdot\vec{c}$ and $|\vec{c}|^2$. $\vec{b}\cdot\vec{c}=-3|\vec{b}|^2=-48$. $|\vec{c}|^2=4|\vec{a}\times\vec{b}|^2+9|\vec{b}|^2=4(16-4)+144=192$. Then $\cos\alpha=\dfrac{-48}{4\sqrt{192}}$ and find $\sin^2\alpha$.
$|\vec{c}|^2=192$. $\vec{b}\cdot\vec{c}=-48$. $\cos^2\alpha=\dfrac{48^2}{16\times192}=\dfrac{3}{4}$. $\sin^2\alpha=\dfrac{1}{4}$. $192\sin^2\alpha=48$.
Correct Answer: 48