Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>If <strong>a</strong> and <strong>b</strong> are two unit vectors, i.e., |<strong>a</strong>| = |<strong>b</strong>| = 1 and angle between them is \(\frac{\pi}{3}\), then \(16[\mathbf{a} \mathbf{b} + \mathbf{a} \times \mathbf{b} \mathbf{b}]\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the scalar triple product property and the cross product magnitude formula to evaluate the expression.
Given: | a | = | b | = 1, angle between them = \(\frac{\pi}{3}\) Step 1: Calculate \(|\mathbf{a} \times \mathbf{b}|\) \(\sin \theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|} \Rightarrow \sin\frac{\pi}{3} = |\mathbf{a} \times \mathbf{b}| = \frac{\sqrt{3}}{2}\) Step 2: Compute scalar triple product \([\mathbf{a} \mathbf{b} + \mathbf{a} \times \mathbf{b} \mathbf{b}] = [\mathbf{a} \mathbf{b} \mathbf{b}] + [\mathbf{a} \mathbf{a} \times \mathbf{b} \mathbf{b}]\) \(= 0 + (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{b} \times \mathbf{a}) = -(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = -|\mathbf{a} \times \mathbf{b}|^2\) \(= -\frac{3}{4}\) Step 3: Final calculation \(16 \times (-\frac{3}{4}) = -12\)
Correct Answer: -12