<p>Let <strong>a</strong> = \(\alpha \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), <strong>b</strong> = \(\mathbf{i} + 2\alpha \mathbf{j} - 2\mathbf{k}\) and <strong>c</strong> = \(2\mathbf{i} - \alpha \mathbf{j} + \mathbf{k}\). Then the value of \(\alpha\) such that \(\{(\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c})\} \times (\mathbf{c} \times \mathbf{a}) = 0\), is</p>
Step-by-Step Solution
Key Concept: For the given vector expression to equal zero, the vectors must be coplanar. This requires the scalar triple product of a, b, c to be zero.
Step 1: Given a = \(\alpha \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), b = \(\mathbf{i} + 2\alpha \mathbf{j} - 2\mathbf{k}\), c = \(2\mathbf{i} - \alpha \mathbf{j} + \mathbf{k}\) Step 2: Use the vector triple product identity: \((\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}) = [\mathbf{a}\mathbf{b}\mathbf{c}]\mathbf{b} - [\mathbf{a}\mathbf{b}\mathbf{b}]\mathbf{c}\) Step 3: Since \([\mathbf{a}\mathbf{b}\mathbf{b}] = 0\) (scalar triple product with repeated vector): \[(\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c}) = [\mathbf{a}\mathbf{b}\mathbf{c}]\mathbf{b}\] Step 4: For the condition \(\{(\mathbf{a} \times \mathbf{b}) \times (\mathbf{b} \times \mathbf{c})\} \times (\mathbf{c} \times \mathbf{a}) = 0\), the vectors must be coplanar or one must be zero. Step 5: This requires that a , b , c are coplanar, i.e., \([\mathbf{a}\mathbf{b}\mathbf{c}] = 0\) Step 6: Compute the determinant: \[\begin{vmatrix} \alpha & 2 & -3 \\ 1 & 2\alpha & -2 \\ 2 & -\alpha & 1 \end{vmatrix} = 0\] Step 7: Expanding and solving yields \(\alpha = 4\) ∴ Answer is 4
Correct Answer: 4