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: Component picture:
Split $\bar a$ into the part along $\bar b$ and the part perpendicular to it. The first has length $a\cos\theta$ and the second $a\sin\theta$.

Step 2: Multiply by $b$:
$|\bar a\cdot\bar b| = b\cdot a\cos\theta$ and $|\bar a\times\bar b| = b\cdot a\sin\theta$.

Step 3: Pythagoras:
Squares of the two perpendicular components of $\bar a$ add to $a^2$, so the two squares times $b^2$ add to $a^2b^2$.

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