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/ Take derivative and set to 0:
Take derivative and set to 0:
February 22, 2026
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\|\vec{d}(t)\|^2 = (2 - 2t)^2 + (-1 - t)^2 + (5 + t)^2
Add them up:
\|\vec{d}(t)\|^2 = (4 - 8t + 4t^2) + (1 + 2t + t^2) + (25 + 10t + t^2) = 6t^2 + 4t + 30
\frac{d}{dt}(6t^2 + 4t + 30) = 12t + 4 = 0 \Rightarrow t = -\frac{1}{3}
Substitute $t = -\frac{1}{3}$ into $\vec{r}(t)$:
\vec{r}\left(-\frac{1}{3}\right) = \begin{pmatrix} 1 + 2(-\frac{1}{3}) \\ -\frac{1}{3} \\ -3 - (-\frac{1}{3}) \end{pmatrix} = \begin{pmatrix} 1 - \frac{2}{3} \\ -\frac{1}{3} \\ -3 + \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ -\frac{8}{3} \end{pmatrix}
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