Question: Find the vector $\mathbf{v}$ that satisfies $\mathbf{v} imes \mathbf{a} = \mathbf{b}$, where $\mathbf{a} = \langle 1, 2, 3
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- ight) + 1 = rac{16}{5} + rac{5}{5} = rac{21}{5}$.
- Thus, the closest point is $\left( rac{8}{5}, rac{21}{5}
- oxed{\left( \dfrac{8}{5},\ \dfrac{21}{5}
- angle$ and $\mathbf{b} = \langle 4, 5, 6
- Solution: The vector equation $\mathbf{v} imes \mathbf{a} = \mathbf{b}$ has a solution only if $\mathbf{a} \cdot \mathbf{b} = 0$, since the cross product is perpendicular to $\mathbf{a}$.
- \mathbf{a} \cdot \mathbf{b} = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32
