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Home » \boxed3x^4 - 12x^2 + 7 - Baxtercollege

\boxed3x^4 - 12x^2 + 7 - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 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}

📌 Thus, the closest point is:

📌 \boxed{\begin{pmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ -\frac{8}{3} \end{pmatrix}}

📌 Question: Find the value of $x$ such that the vectors $\begin{pmatrix} 2 \\ x \\ -3 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$ are orthogonal.

📌 Two vectors are orthogonal if their dot product is zero. Compute the dot product:

📌 \begin{pmatrix} 2 \\ x \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} = 2 \cdot 4 + x \cdot (-1) + (-3) \cdot 1 = 8 - x - 3 = 5 - x

📌 - x = 0 \Rightarrow x = 5

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