Let $vecy$ = $ left< eginarraycc 8 \ 5 \ -5 \ endarray
ight>$ and $vecu$ = $ left< eginarraycc 4 \ 4 \ 4 \ endarray
ight>$. Compute the distance $d$ from $vecy$ to the line with $vecu$ and also the beginning.

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$$d=left|vecy-fracvecycdotvecuvecu ight|$$Then$$d=left|eginpmatrix8\5\-5endpmatrix-frac324sqrt3eginpmatrix4\4\4endpmatrix ight|$$

I"ll leave the remainder to you

The squared distance from $(8,5,-5)$ to a point $(4t,4t,4t)$ on the line is$$(8-4t)^2+(5-4t)^2+(-5-4t)^2=48t^2-64t+114. ag1$$This is minimal at $t=64/96=2/3$ so that the nearest suggest on the line is $(8/3,8/3,8/3).$Or because you only require the distance, plug $t=2/3$ right into (1) and also acquire $278/3$, and the distance is the squareroot of that.

NOTE: The line might be more ssuggest $(t,t,t)$ which would certainly make for smaller coefficients in (1).

You can think of $vecy$ as being the amount of 2 vectors: a vector $vecy_p$ parallel to the line via $vecu$ and the origin, and also a vector $vecy_o$ orthogonal to that line.

The $d$ you should compute is the length of $vecy_o$.

$vecy_p$ is the estimate of $vecy$ onto the line with $vecu$ and the beginning. The length of $vecy_p$ is equal to the absolute value of the dot product $vecycdotvecu"$ where $vecu"$ is a unit vector in the very same direction as $vecu$.

Knowing the size of $vecy$ and the length of $vecy_p$ you deserve to use the Pythagorean Theorem to gain the size of $vecy_o$.

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