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D = (x - 4)^2 + (2x + 1 - 3)^2 = (x - 4)^2 + (2x - 2)^2.
D = (x^2 - 8x + 16) + (4x^2 - 8x + 4) = 5x^2 - 16x + 20.
rac{dD}{dx} = 10x - 16 = 0 \Rightarrow x = rac{8}{5}.
Then $y = 2\left( rac{8}{5}
ight) + 1 = rac{16}{5} + rac{5}{5} = rac{21}{5}$.
Thus, the closest point is $\left( rac{8}{5}, rac{21}{5}
oxed{\left( \dfrac{8}{5},\ \dfrac{21}{5}
Question: Find the vector $\mathbf{v}$ that satisfies $\mathbf{v} imes \mathbf{a} = \mathbf{b}$, where $\mathbf{a} = \langle 1, 2, 3
angle$ 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 $\mathbf{a}$.