Vector Algebra
Cross Product and Scalar Product
Grade None
Question:
<p><strong>a</strong> = \(2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), <strong>b</strong> = \(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\), <strong>c</strong> = \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) and <strong>d</strong> = \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), then \(\frac{1}{7}(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d})\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the vector identity for mixed product to convert the cross product dot product into simpler scalar products.
Given: a = \(2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), b = \(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\), c = \(\mathbf{i} + \mathbf{j} + \mathbf{k}\), d = \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) Step 1: Use the identity \((\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d}) - (\mathbf{b} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{d})\) Step 2: Calculate dot products \(\mathbf{a} \cdot \mathbf{c} = 2(1) + 3(1) + (-1)(1) = 4\) \(\mathbf{b} \cdot \mathbf{d} = (-1)(3) + 2(2) + (-4)(1) = -3\) \(\mathbf{b} \cdot \mathbf{c} = (-1)(1) + 2(1) + (-4)(1) = -3\) \(\mathbf{a} \cdot \mathbf{d} = 2(3) + 3(2) + (-1)(1) = 11\) Step 3: Substitute \((4)(-3) - (-3)(11) = -12 + 33 = 21\) Step 4: Final answer \(\frac{1}{7} \times 21 = 3\)
Correct Answer: 3