Question

Let a and b be two non-collinear vectors of unit modulus. If u = a − (a · b)b and v = a × b, then ∥v∥ = ?

• A. \( \lVert \mathbf{u} \rVert + \lVert \mathbf{u} \cdot \mathbf{v} \rVert \)
• B. \( \frac{\lVert \mathbf{u} \rVert}{2} \)
• C. \( \lVert \mathbf{u} \rVert + \frac{\lVert \mathbf{u} \cdot \mathbf{b} \rVert}{2} \)
• D. \( \frac{\lVert \mathbf{u} \rVert}{5} \)

Hint:

For unit vectors, the cross product’s magnitude is determined by the sine of the angle between them. For non-collinear vectors, the sine value is 1.

Answer 1

The Correct Option is A

Step 1: We are given: \[ \mathbf{u} = \mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{b} \quad \text{and} \quad \mathbf{v} = \mathbf{a} \times \mathbf{b} \] Where: - \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors (i.e., \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 1 \)). 

Step 1: Compute \( \lVert \mathbf{u} \rVert \) 
Using the identity for vector projection, \[ \mathbf{u} = \mathbf{a} - \text{Proj}_{\mathbf{b}} \mathbf{a} \] The projection formula is: \[ \text{Proj}_{\mathbf{b}} \mathbf{a} = (\mathbf{a} \cdot \mathbf{b}) \mathbf{b} \] Since \( \mathbf{u} \) is the component of \( \mathbf{a} \) perpendicular to \( \mathbf{b} \), we can compute its magnitude: \[ \lVert \mathbf{u} \rVert = \sqrt{|\mathbf{a}|^2 - (\mathbf{a} \cdot \mathbf{b})^2} \] Since \( |\mathbf{a}| = 1 \), \[ \lVert \mathbf{u} \rVert = \sqrt{1 - (\cos \theta)^2} = \sqrt{\sin^2 \theta} = |\sin \theta| \] 

Step 2: Compute \( \lVert \mathbf{v} \rVert \) 
Recall that \( \mathbf{v} = \mathbf{a} \times \mathbf{b} \). By the cross product formula: \[ \lVert \mathbf{v} \rVert = |\mathbf{a}| |\mathbf{b}| \sin \theta = 1 \cdot 1 \cdot |\sin \theta| = |\sin \theta| \] Thus, \[ \lVert \mathbf{v} \rVert = \lVert \mathbf{u} \rVert \] 

Step 3: Relating \( \lVert \mathbf{v} \rVert \) to Other Terms 
Since \( \mathbf{v} = \mathbf{a} \times \mathbf{b} \), and the cross product is perpendicular to both vectors, \[ \lVert \mathbf{v} \rVert = \lVert \mathbf{u} \rVert + \lVert \mathbf{u} \cdot \mathbf{v} \rVert \] 

Step 4: Final Answer 
\[ \boxed{\lVert \mathbf{u} \rVert + \lVert \mathbf{u} \cdot \mathbf{v} \rVert} \] 

Final Answer: (A) \( \lVert \mathbf{u} \rVert + \lVert \mathbf{u} \cdot \mathbf{v} \rVert \)