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Let $L$ be the line in $\Bbb R^3$ containing the point $A (0,1,1)$.

Specifically, let $L = \{t[0,0,1]+[0,1,1]\mid t\in\Bbb R\}$.

Let $U=[1,-1,0]$

The question is to find the point on the line which is closet to the point $U$.

And here is the answer:

Let $P$ be the point we are looking for.

Then $AU =[1,-2,-1]$.

We see that $AP=\operatorname{proj}_n (AU)$, where $n$ is the normal vector of the line which is $[0,0,1]$.

My question is why $\operatorname{proj}_n (AU)$ equals $AP$. I think it should be $UP$.

dfeuer
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Jing
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1 Answers1

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In the first sentence, there are many lines containing $A$, but that is fixed in the next sentence. In the next to last sentence, there is not normal vector to $L$. There is a direction vector, which is what you quote.

When you project $AU$ onto $[0,0,1]$ it can't be $UP$ as $P$ is on $L$ and $U$ is not. What you are finding is the distance to move from $A$ along the line until the vector $UP$ is perpendicular to the line.

Ross Millikan
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