Let a = 3 i + j - 2 k, b = 4 i + j + 7 k$\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}, \vec{b} = 4\hat{i} + \hat{j} + 7\hat{k}$ and c = i - 3 j + 4 k$\vec{c} = \hat{i} - 3\hat{j} + 4\hat{k}$ be three vectors. If a vector p$\vec{p}$ satisfies p × b = c × b$\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$ and p · a = 0$\vec{p} \cdot \vec{a} = 0$, then p · ( i - j - k)$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ 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