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

Answer: (A)

90

Answer 2

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