Using the identity $ \mathbf{a} \times (\mathbf{v} \times \mathbf{a}) = (\mathbf{a} \cdot \mathbf{a})\mathbf{v} - (\mathbf{a} \cdot \mathbf{v})\mathbf{a} $, and since $ \mathbf{a} \cdot \mathbf{v} = 0 $, this becomes:

February 22, 2026

$Using the identity $ \mathbf{a} \times (\mathbf{v} \times \mathbf{a}) = (\mathbf{a} \cdot \mathbf{a})\mathbf{v} - (\mathbf{a} \cdot \mathbf{v})\mathbf{a} $, and since $ \mathbf{a} \cdot \mathbf{v} = 0 $, this becomes:$

\mathbf{v} \times \mathbf{a} = \mathbf{b}, \quad \mathbf{v} \cdot \mathbf{a} = 0
We use the vector triple product identity. From $ \mathbf{v} \times \mathbf{a} = \mathbf{b} $, taking the cross product with $ \mathbf{a} $ on both sides:
\mathbf{a} \times (\mathbf{v} \times \mathbf{a}) = \mathbf{a} \times \mathbf{b}
\|\mathbf{a}\|^2 \mathbf{v} = \mathbf{a} \times \mathbf{b}
Compute $ \|\mathbf{a}\|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 $
Now compute $ \mathbf{a} \times \mathbf{b} $:

Using the identity $ \mathbf{a} \times (\mathbf{v} \times \mathbf{a}) = (\mathbf{a} \cdot \mathbf{a})\mathbf{v} - (\mathbf{a} \cdot \mathbf{v})\mathbf{a} $, and since $ \mathbf{a} \cdot \mathbf{v} = 0 $, this becomes:

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