Let a$\vec{a}$ and b$\vec{b}$ be two vectors such that | a| = 1, | b| = 4$|\vec{a}| = 1, |\vec{b}| = 4$ and a · b = 2$\vec{a} \cdot \vec{b} = 2$. If c = (2 a × b) - 3 b$\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between b$\vec{b}$ and c$\vec{c}$ is α$\alpha$, then 192 ²α$192\sin^2\alpha$ is equal to

Core Logic

Since b · ( a × b) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$, we have |b||c| α = -3|b|²$|b||c|\cos\alpha = -3|b|^2$.

Step 1: Compute Modulus of c

Given a · b = 2 |a||b| θ = 2 1 · 4 θ = 2 θ = (π)/(3)$\vec{a} \cdot \vec{b} = 2 \implies |a||b|\cos\theta = 2 \implies 1 \cdot 4 \cos\theta = 2 \implies \theta = \frac{\pi}{3}$.

Step 2: Final Calculation

We know |c|² ² α = 144$|c|^2 \cos^2 \alpha = 144$.

Chapter Mix

Class 12 Maths: Vector Algebra