Consider two vectors u = 3 i - j$\vec{u} = 3\hat{i} - \hat{j}$ and v = 2 i + j - λ k$\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, where λ > 0$\lambda > 0$. The angle between them is given by ⁻¹( √(5)2sqrt7)$\cos^{-1}\left(\frac{\sqrt{5}}{2sqrt{7}}\right)$. Let v = v₁ + v₂$\vec{v} = \vec{v}_1 + \vec{v}_2$, where v₁$\vec{v}_1$ is parallel to u$\vec{u}$ and v₂$\vec{v}_2$ is perpendicular to u$\vec{u}$. Then the value | v₁|² + | v₂|²$|\vec{v}_1|^2 + |\vec{v}_2|^2$ 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