Question

The value of \(\{(\vec{a} \times \vec{b})^2 + (\vec{a}\cdot \vec{b})^2\\div a^2 b^2\) is}

• A. 0
• B. 1
• C. 2
• D. None of these

Hint:

Always remember: \(|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = a^2 b^2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses Lagrange's Identity, which relates the magnitudes of the dot product and cross product of two vectors to their individual magnitudes.
Step 2: Detailed Explanation:
Let the angle between \( \vec{a} \) and \( \vec{b} \) be \( \theta \).
1. Magnitude of dot product: \( (\vec{a} \cdot \vec{b})^{2} = (|\vec{a}||\vec{b}| \cos \theta)^{2} = a^{2} b^{2} \cos^{2} \theta \).
2. Magnitude of cross product: \( (\vec{a} \times \vec{b})^{2} = (|\vec{a}||\vec{b}| \sin \theta)^{2} = a^{2} b^{2} \sin^{2} \theta \).
Now, sum them: \[ (\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = a^{2} b^{2} \sin^{2} \theta + a^{2} b^{2} \cos^{2} \theta \] Factor out \( a^{2} b^{2} \): \[ = a^{2} b^{2} (\sin^{2} \theta + \cos^{2} \theta) \] Using the trigonometric identity \( \sin^{2} \theta + \cos^{2} \theta = 1 \): \[ = a^{2} b^{2} \].
The given expression is \( \frac{a^{2} b^{2}}{a^{2} b^{2}} = 1 \).
Step 3: Final Answer:
The value is 1.