Question

The value of \(|\overset{̄}{a}\cdot \overset{̄}{b}|^2+|\overset{̄}{a}\times \overset{̄}{b}|\cdot |\overset{̄}{a}\times \overset{̄}{b}|\) is \(\ldots\)

• A. \(-a^2b^2\)
• B. \(a^2b^2\)
• C. \(a^2b^2cosθ\)
• D. \(a^2b^2sinθ\)

Hint:

Use a.b = ab cos(theta) and |a x b| = ab sin(theta), then add.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For two vectors with magnitudes \(a\) and \(b\) and angle \(\theta\) between them, the dot product and the magnitude of the cross product are given by simple formulas.

Step 2: Key Formula or Approach:
1. \(\bar a\cdot\bar b = ab\cos\theta\).
2. \(|\bar a\times\bar b| = ab\sin\theta\).

Step 3: Detailed Explanation:
\[ |\bar a\cdot\bar b|^2 = a^2b^2\cos^2\theta \]
\[ |\bar a\times\bar b|\cdot|\bar a\times\bar b| = a^2b^2\sin^2\theta \]
Add:
\[ a^2b^2\left(\cos^2\theta + \sin^2\theta\right) = a^2b^2 \]
This is Lagrange's identity. Options (A), (C) and (D) either have the wrong sign or carry a leftover trigonometric factor, which cannot be since \(\cos^2\theta + \sin^2\theta = 1\).

Final Answer:
The value is \(a^2b^2\), option (B). \[ \boxed{a^2b^2 \text{ (B)}} \]