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Home » The general form is \( y = ax^2 \). - DNSFLEX

The general form is \( y = ax^2 \). - DNSFLEX

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Let the point on the line be $\vec{r}(t) = \begin{pmatrix} 1 + 2t \\ t...
Let $P = (3, -1, 2)$. The vector from the point on the line to $P$ is:
\vec{d}(t) = \vec{P} - \vec{r}(t) = \begin{pmatrix} 3 - (1 + 2t) \\ -1...
To find the closest point, minimize the magnitude of $\vec{d}(t)$, or...
\|\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)...
Take derivative and set to 0:
\frac{d}{dt}(6t^2 + 4t + 30) = 12t + 4 = 0 \Rightarrow t = -\frac{1}{3...
Substitute $t = -\frac{1}{3}$ into $\vec{r}(t)$:

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