I"m having a little challenge acquiring this trouble down. I"ve been trying to follow my notes, yet I guess I"m not doing it correctly. Anyone know how to appropriately answer this question?

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-frac324sqrt3eginpmatrix4\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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