Vector Algebra
Cross Product and Linear Constraints
Grade 12
Question:
<p>If \(\mathbf{a} + 2\mathbf{b} + 3\mathbf{c} = \mathbf{0}\), then \(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} =\)</p><p>(a) \(2(\mathbf{a} \times \mathbf{b})\)</p><p>(b) \(6(\mathbf{b} \times \mathbf{c})\)</p><p>(c) \(3(\mathbf{c} \times \mathbf{a})\)</p><p>(d) \(\mathbf{0}\)</p>
<p>(a) \(2(\mathbf{a} \times \mathbf{b})\)</p>
<p>(b) \(6(\mathbf{b} \times \mathbf{c})\)</p>
<p>(c) \(3(\mathbf{c} \times \mathbf{a})\)</p>
<p>(d) \(\mathbf{0}\)</p>
Step-by-Step Solution
Key Concept: Use the linear constraint to express one vector in terms of others and simplify using properties of cross product: \(\mathbf{u} \times \mathbf{u} = \mathbf{0}\) and \(\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}\).
Given: \(\mathbf{a} + 2\mathbf{b} + 3\mathbf{c} = \mathbf{0}\) Therefore: \(\mathbf{a} = -(2\mathbf{b} + 3\mathbf{c})\) Now, \(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}\) \(= -(2\mathbf{b} + 3\mathbf{c}) \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \{-(2\mathbf{b} + 3\mathbf{c})\}\) \(= -3\mathbf{c} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} - 2\mathbf{c} \times \mathbf{b}\) \(= 3\mathbf{b} \times \mathbf{c} + \mathbf{b} \times \mathbf{c} + 2\mathbf{b} \times \mathbf{c}\) \(= 6(\mathbf{b} \times \mathbf{c})\) Similarly, by expressing other vectors in terms of the constraint, we can verify options (a) and (c) are also correct. ∴ Correct answers are (a, b, c)
Correct Answer: A, B, C