Question

Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors. Let \( \theta \) be the angle between \( \vec{a} \) and \( \vec{b} \). If \( \theta \neq 0 \) or \( \pi \), then

\[ \left| \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \right|^2 \] is equal to:

• A. \( \cos^2 \theta \)
• B. \( \sin^2 \theta \)
• C. \( \tan^2 \theta \)
• D. \( 1 \)
• E. \( 2 \cos^2 \theta \)

Hint:

For unit vectors \( \vec{a} \) and \( \vec{b} \), the dot product \( \vec{a} \cdot \vec{b} = \cos \theta \), and this identity simplifies many vector-related expressions.

Answer 1

The Correct Option is B

Step 1: The given expression is: \[ |\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}|^2. \] First, expand the square of the vector: \[ |\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}|^2 = (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}). \] Using the distributive property of the dot product: \[ = \vec{a} \cdot \vec{a} - 2 (\vec{a} \cdot \vec{b}) (\vec{a} \cdot \vec{b}) + (\vec{a} \cdot \vec{b})^2 \vec{b} \cdot \vec{b}. \] Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have \( \vec{a} \cdot \vec{a} = 1 \) and \( \vec{b} \cdot \vec{b} = 1 \). 
So the expression becomes: \[ 1 - 2 (\vec{a} \cdot \vec{b})^2 + (\vec{a} \cdot \vec{b})^2. \] Thus, we get: \[ 1 - (\vec{a} \cdot \vec{b})^2. \] Now, using the formula for the dot product of two unit vectors: \[ \vec{a} \cdot \vec{b} = \cos \theta. \] Therefore, the expression becomes: \[ 1 - \cos^2 \theta. \] Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can simplify this to: \[ \sin^2 \theta. \] 
Thus, the correct answer is option (B).