I"m having a little an obstacle getting this difficulty down. I"ve to be trying to monitor my notes, yet I guess I"m not doing the correctly. Anyone know just how to effectively answer this question?

Let $\vecy$ = $ \left< \beginarraycc 8 \\ 5 \\ -5 \\ \endarray \right>$ and also $\vecu$ = $ \left< \beginarraycc 4 \\ 4 \\ 4 \\ \endarray \right>$. Compute the distance $d$ from $\vecy$ to the line through $\vecu$ and the origin.

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I"ll leaving the rest to you


You have the right to think that $\vecy$ as being the sum of two vectors: a vector $\vecy_p$ parallel come the line v $\vecu$ and also the origin, and also a vector $\vecy_o$ orthogonal to the line.

The $d$ you have to compute is the size of $\vecy_o$.

$\vecy_p$ is the forecast of $\vecy$ onto the line with $\vecu$ and also the origin. The length of $\vecy_p$ is equal to the absolute value of the period product $\vecy\cdot\vecu"$ whereby $\vecu"$ is a unit vector in the same direction as $\vecu$.

Knowing the length of $\vecy$ and also the length of $\vecy_p$ you have the right to use the Pythagorean organize to obtain the length of $\vecy_o$.



The squared street from $(8,5,-5)$ come a suggest $(4t,4t,4t)$ ~ above the line is$$(8-4t)^2+(5-4t)^2+(-5-4t)^2=48t^2-64t+114.\tag1$$This is minimal at $t=64/96=2/3$ so the the nearest allude on the heat is $(8/3,8/3,8/3).$Or since you only need the distance, plug $t=2/3$ into (1) and also get $278/3$, and the street is the squareroot of that.

NOTE: The line can be much more simply $(t,t,t)$ which would make for smaller sized coefficients in (1).


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