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Home » Using the quadratic formula: \(x = \frac-b \pm \sqrtb^2 - 4ac2a\). - Dachbleche24

Using the quadratic formula: \(x = \frac-b \pm \sqrtb^2 - 4ac2a\). - Dachbleche24

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Let the point on the line be $\vec{r}(t) = \begin{pmatrix} 1 + 2t \\ t \\ -3 - t \end{pmatrix}$.

📌 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 - t \\ 2 - (-3 - t) \end{pmatrix} = \begin{pmatrix} 2 - 2t \\ -1 - t \\ 5 + t \end{pmatrix}

📌 To find the closest point, minimize the magnitude of $\vec{d}(t)$, or equivalently, minimize $\|\vec{d}(t)\|^2$:

📌 \|\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

📌 Take derivative and set to 0:

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