Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>If |<strong>a</strong>| = |<strong>b</strong>| = |<strong>c</strong>| = 2 and \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{a} = 2\), then \([\mathbf{a} \mathbf{b} \mathbf{c}] \cos 45°\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the Gram determinant to find the scalar triple product, then multiply by the given cosine factor.
Given: | a | = | b | = | c | = 2, \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{a} = 2\) Step 1: Use Gram determinant formula \([\mathbf{a} \mathbf{b} \mathbf{c}]^2 = \begin{vmatrix} \mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{vmatrix} = \begin{vmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{vmatrix}\) Step 2: Calculate determinant \(= 4(16-4) - 2(8-4) + 2(4-8) = 48 - 8 - 8 = 32\) Step 3: Find [ a b c ] \([\mathbf{a} \mathbf{b} \mathbf{c}] = \sqrt{32} = 4\sqrt{2}\) Step 4: Apply cosine \(4\sqrt{2} \cdot \cos 45° = 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 4\)
Correct Answer: 4