Question

For two non-zero, non-collinear, non-orthogonal vectors \(\vec{a}\) and \(\vec{b}\) where \(\theta\) is angle between them, then \(\frac{(\vec{a}\cdot \vec{b})^{2}+|\vec{a}\times \vec{b}|^{2}}{(\vec{a}\cdot \vec{b})^{2}-|\vec{a}\times \vec{b}|^{2}}=\)

• A. \(\tan \left(\frac{\pi }{4}+\theta \right)\)
• B. \(\tan \left(\frac{\pi }{4}-\theta \right)\)
• C. \(\sec 2\theta\)
• D. \(\csc 2\theta\)

Hint:

Write the dot product with cos and the cross product magnitude with sin, then use the double angle identity.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Use \(\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\) and \(|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta\).

Step 2: Substitute
The expression becomes \(\frac{|\vec{a}|^{2}|\vec{b}|^{2}(\cos^{2}\theta+\sin^{2}\theta)}{|\vec{a}|^{2}|\vec{b}|^{2}(\cos^{2}\theta-\sin^{2}\theta)}\).

Step 3: Simplify
The common factor cancels, leaving \(\frac{1}{\cos 2\theta}=\sec 2\theta\).

Final Answer:
The expression equals \(\sec 2\theta\). \[ \boxed{\text{(C) }\sec 2\theta} \]