Let a vector a = √(2) i - j + λ k$\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}$, λ > 0$\lambda > 0$, make an obtuse angle with the vector b = -λ² i + 4√(2) j + 4√(2) k$\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle θ$\theta$, (π)/(6) < θ < (π)/(2)$\frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of λ$\lambda$ is (α, β) - γ$(\alpha, \beta) - \{\gamma\}$, then α + β + γ$\alpha + \beta + \gamma$ 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