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

Answer 2

To solve the problem, let's break it down step-by-step using vector algebra and trigonometry principles.

1. \(\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}\) is given, where \(\alpha, \beta \in \mathbb{R}\).
2. The vector \(\vec{b}\) has the property that the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is given as \(\theta = \frac{\pi}{4}\). We also know that \(|\vec{b}|^2 = 6\).
3. The magnitude of the cross product \(|\vec{a} \times \vec{b}|\) is given as \(3\sqrt{2}\).
4. We know that the magnitude of the cross product can also be calculated using: \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\). Given that \(\theta = \frac{\pi}{4}\), \(\sin\theta = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\).
5. Therefore, the equation becomes: \(3\sqrt{2} = |\vec{a}||\vec{b}| \cdot \frac{\sqrt{2}}{2}\).
   Simplifying this gives: \(|\vec{a}||\vec{b}| = 6\).
6. Since \(|\vec{b}|^2 = 6\), we find \(|\vec{b}| = \sqrt{6}\). Now calculate \(|\vec{a}|\):
   \(|\vec{a}| = \frac{6}{\sqrt{6}} = \sqrt{6}\).
7. The magnitude of \(\vec{a}\) can also be represented as: \(|\vec{a}| = \sqrt{1 + \alpha^2 + \beta^2}\).
8. Equating the two expressions for \(|\vec{a}|\), we get: \(\sqrt{1 + \alpha^2 + \beta^2} = \sqrt{6}\).
   Squaring both sides gives: \(1 + \alpha^2 + \beta^2 = 6\), thus: \(\alpha^2 + \beta^2 = 5\).
9. We need to find: \((\alpha^2 + \beta^2) \times |\vec{a} \times \vec{b}|^2\).
   Substituting the known values: \(= 5 \times (3\sqrt{2})^2\)
   \(= 5 \times 18\)
   \(= 90\).

Therefore, the value of \((\alpha^2 + \beta^2) |\vec{a} \times \vec{b}|^2\) is 90.