Vector Algebra
Scalar Triple Product and Coplanar Vectors
Grade 12
Question:
<p>If \(\alpha(\mathbf{a} \times \mathbf{b}) + \beta(\mathbf{b} \times \mathbf{c}) + \gamma(\mathbf{c} \times \mathbf{a}) = \mathbf{0}\), then</p><p>(a) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are coplanar if all of \(\alpha, \beta, \gamma \neq 0\)</p><p>(b) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are coplanar if any one of \(\alpha, \beta, \gamma \neq 0\)</p><p>(c) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are non-coplanar for any \(\alpha, \beta, \gamma \neq 0\)</p><p>(d) None of the above</p>
<p>(a) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are coplanar if all of \(\alpha, \beta, \gamma \neq 0\)</p>
<p>(b) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are coplanar if any one of \(\alpha, \beta, \gamma \neq 0\)</p>
<p>(c) \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are non-coplanar for any \(\alpha, \beta, \gamma \neq 0\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: The scalar triple product \([\mathbf{a} \mathbf{b} \mathbf{c}] = 0\) if and only if vectors are coplanar. Taking dot products with the given equation reveals that any non-zero coefficient forces the triple product to zero.
Given: \(\alpha(\mathbf{a} \times \mathbf{b}) + \beta(\mathbf{b} \times \mathbf{c}) + \gamma(\mathbf{c} \times \mathbf{a}) = \mathbf{0}\) Taking dot product with \(\mathbf{c}\): \(\alpha[\mathbf{a} \mathbf{b} \mathbf{c}] + \beta[\mathbf{b} \mathbf{c} \mathbf{c}] + \gamma[\mathbf{c} \mathbf{c} \mathbf{a}] = 0\) \(\Rightarrow \alpha[\mathbf{a} \mathbf{b} \mathbf{c}] = 0\) Similarly, taking dot product with \(\mathbf{b}\) and \(\mathbf{a}\): \(\gamma[\mathbf{a} \mathbf{b} \mathbf{c}] = 0\) and \(\beta[\mathbf{a} \mathbf{b} \mathbf{c}] = 0\) Now, even if any one of \(\alpha, \beta, \gamma \neq 0\), then we must have \([\mathbf{a} \mathbf{b} \mathbf{c}] = 0\) \(\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}\) are coplanar. ∴ Correct answers are (a, b)
Correct Answer: A, B