Question

Math

Posted 5 months ago

`Problem: Let $\mathbf{a}=\left(\begin{array}{c}5 \\ -3 \\ -4\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}-11 \\ 1 \\ 28\end{array}\right)$. 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 $\mathbf{d}$.`

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Answer from Sia

Posted 5 months ago

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: $\mathbf{d} = \mathbf{b} - \mathbf{a} = \left(\begin{array}{c}-11 - 5 \\ 1 - (-3) \\ 28 - (-4)\end{array}\right)$

step 3

Calculating the components of $\mathbf{d}$: $\mathbf{d} = \left(\begin{array}{c}-16 \\ 4 \\ 32\end{array}\right)$

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 $\frac{\mathbf{d}}{\|\mathbf{d}\|} = \left(\begin{array}{c}-16/36 \\ 4/36 \\ 32/36\end{array}\right)$

step 7

Simplifying the components of the normalized vector $\mathbf{d}$: $\mathbf{d} = \left(\begin{array}{c}-4/9 \\ 1/9 \\ 8/9\end{array}\right)$

Answer

$\mathbf{d} = \left(\begin{array}{c}-4/9 \\ 1/9 \\ 8/9\end{array}\right)$

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 $t$.

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