Problem: Let $\mathbf{a}=\left(\begin{array}{c}5 \ -3 \ -4\end{array} ight)$ and $\mathbf{b}=\left(\begin{array}{c}-11 \ 1 \ 28\end{array} ight)$. There exist vectors $\mathbf{p}$ and $\mathbf{d}$ such that the line containing $\mathbf{a}$ and $\mathbf{b}$ can be expressed in the form $\mathbf{v}=\mathbf{p}+\mathbf{d} t$. Furthermore, for a certain choice of $\mathbf{d}$, it is the case that for all points $\mathbf{v}$ lying on the same side of $\mathbf{a}$ that $\mathbf{b}$ lies on, the distance between $\mathbf{v}$ and $\mathbf{a}$ is $t$. Find $\mathbf{d}$.

Question

Problem: Let $\mathbf{a}=\left(\begin{array}{c}5 \ -3 \ -4\end{array}ight)$ and $\mathbf{b}=\left(\begin{array}{c}-11 \ 1 \ 28\end{array}ight)$. There exist vectors $\mathbf{p}$ and $\mathbf{d}$ such that the line containing $\mathbf{a}$ and $\mathbf{b}$ can be expressed in the form $\mathbf{v}=\mathbf{p}+\mathbf{d} t$. Furthermore, for a certain choice of $\mathbf{d}$, it is the case that for all points $\mathbf{v}$ lying on the same side of $\mathbf{a}$ that $\mathbf{b}$ lies on, the distance between $\mathbf{v}$ and $\mathbf{a}$ is $t$. Find $\mathbf{d}$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the vector $\mathbf{d}$, we subtract vector $\mathbf{a}$ from vector $\mathbf{b}$. ‖ ‖ step 2 ⋮ The subtraction is performed component-wise: $\mathbf{d} = \mathbf{b} - \mathbf{a} = \left(\begin{array}{c}-11 - 5 \ 1 - (-3) \ 28 - (-4)\end{array} ight)$. ‖ ‖ step 3 ⋮ Calculating the components of $\mathbf{d}$: $\mathbf{d} = \left(\begin{array}{c}-16 \ 4 \ 32\end{array} ight)$. ‖ ‖ step 4 ⋮ To normalize vector $\mathbf{d}$, we divide it by its magnitude. The magnitude of $\mathbf{d}$ is $\|\mathbf{d}\| = \sqrt{(-16)^2 + 4^2 + 32^2}$. ‖ ‖ step 5 ⋮ Calculating the magnitude: $\|\mathbf{d}\| = \sqrt{256 + 16 + 1024} = \sqrt{1296} = 36$. ‖ ‖ step 6 ⋮ The normalized vector $\mathbf{d}$ is then $\frac{\mathbf{d}}{\|\mathbf{d}\|} = \left(\begin{array}{c}-16/36 \ 4/36 \ 32/36\end{array} ight)$. ‖ ‖ step 7 ⋮ Simplifying the components of the normalized vector $\mathbf{d}$: $\mathbf{d} = \left(\begin{array}{c}-4/9 \ 1/9 \ 8/9\end{array} ight)$. ‖ ∻Answer∻ ⚹ $\mathbf{d} = \left(\begin{array}{c}-4/9 \ 1/9 \ 8/9\end{array} ight)$ ⚹ ∻Key Concept∻ ⚹Vector Subtraction and Normalization⚹ ∻Explanation∻ ⚹To find a direction vector $\mathbf{d}$ for a line containing points $\mathbf{a}$ and $\mathbf{b}$, we subtract $\mathbf{a}$ from $\mathbf{b}$ and normalize the resulting vector. Normalization is done by dividing the vector by its magnitude to ensure that the distance from $\mathbf{a}$ to any point $\mathbf{v}$ on the line is equal to the scalar multiple $t$.⚹◊What is the equation of the line passing through vectors $\mathbf{a}$ and $\mathbf{b}$ in the form $\mathbf{v} = \mathbf{p} + \mathbf{d} t$?⍭ Generate me a similar question◊

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