Question

Let \( \vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k} \), \( \alpha, \beta \in \mathbb{R} \). Let a vector \( \vec{b} \) be such that the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \) and \( |\vec{b}|^2 = 6 \),
If \( \vec{a} \times\vec{b} = 3\sqrt{2} \), then the value of \( \left( \alpha^2 + \beta^2 \right) | \vec{a} \times \vec{b} |^2 \) is equal to

• A. 90
• B. 75
• C. 95
• D. 85

Answer 1

The Correct Option is A

Using \(\vec{a} \times \vec{b} = |\vec{a}||\vec{b}| \cos \theta\):

\(3\sqrt{2} = |\vec{a}| \times 6 \times \frac{\sqrt{2}}{2} \Rightarrow |\vec{a}| = 1.\)

Since \(|\vec{a}|^2 = 1\), we have \(1 + \alpha^2 + \beta^2 = 1 \Rightarrow \alpha^2 + \beta^2 = 5.\)

For \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta\):

\(|\vec{a} \times \vec{b}| = 1 \times 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2}.\)

Thus, \((\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = 5 \times 18 = 90.\)