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Compute $ \|\mathbf{a}\|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 $Now compute $ \mathbf{a} \times \mathbf{b} $:\mathbf{a} \times \mathbf{b} = \begin{vmatrix}& -2 & 1 \\\end{vmatrix} = \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot (-2) - 2 \cdot 1)= \mathbf{i}(2 + 6) - \mathbf{j}(1 - 3) + \mathbf{k}(-2 - 2) = \langle 8, 2, -4 \rangle14\mathbf{v} = \langle 8, 2, -4 \rangle \quad \Rightarrow \quad \mathbf{v} = \frac{1}{14} \langle 8, 2, -4 \rangle = \left\langle \frac{4}{7}, \frac{1}{7}, -\frac{2}{7} \right\rangleThus, the vector is $ \boxed{\left\langle \dfrac{4}{7}, \dfrac{1}{7}, -\dfrac{2}{7} \right\rangle} $.Question: Among all roots of the polynomial $ z^6 - z^4 + z^2 - 1 = 0 $, a historian of mathematical astronomy seeks the maximum imaginary part, expressed as $ \sin \theta $ for $ 0 < \theta < \pi $. Find $ \theta $.Solution: Consider the polynomial: