Find value of $t$ between the difference of 3D vectors.

Question

Hint: The distance between $2$ vectors equals the magnitude of their difference.

What is the value of $t$ for which the vector $\mathbf v = \begin{pmatrix} 2 \\ -3 \\ -3 \end{pmatrix} + \begin{pmatrix} 7 \\ 5 \\ -1 \end{pmatrix}t$ is closest to $\mathbf a = \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix}?$

Using the hint, my idea is to find the difference of v and a and find the magnitude of the value. From there, I can use regular algebraic forms to solve for $t$.

This is what I have done so far:

$\mathbf v$ can be simplified down to $\begin{pmatrix} 2+7t \\ -3+5t \\ -3-t \end{pmatrix}$.

The difference of $\mathbf v$ and $\mathbf a$ is: $\begin{pmatrix} 2+7t \\ -3+5t \\ -3-t \end{pmatrix} - \begin{pmatrix} 4 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} -2+7t \\ -7+5t \\ -8-t \end{pmatrix}$.

Am I solving this problem correctly so far, and do I have the right idea as to how to properly approach this problem?

Thanks.

--Grace

Answer 1

Ok, it's long for comment. We should minimize $$ |v−a|^2=(7t−2)^2+(5t−7)^2+(−t−8)^2=75t^2-82t+117 $$ (actually, it's distance from point $\vec a$ to the line $\vec v$). Without derivatives we can write $$ 75t^2-82t+117=75\left(t^2-\frac{82}{75}t + \frac{41^2}{75^2}\right) + \left(117 - \frac{41^2}{75}\right) =\\= 75\left(t-\frac{41}{75}\right)^2 + \left(117 - \frac{41^2}{75}\right) \ge 117 - \frac{41^2}{75}$$

Answer 2

For a straight line given by $x=b+\lambda u$ and a point $ a$ consider the area of the parallelogram spanned by $u$ and $a-b$. Its area is the length of the cross product of those two. Divide that area by the length of $u$ to get the height of the parallelogram which is the distance from $a$ to the straight line.