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I would like to understand how to project a point onto a line.

The first method that comes to my mind would be to say that the line is defined by two points $q_1, q_2$. Point $p$ is projected onto that line by finding a point $p'$ on the line that is closest to $p$. the vector $\vec{pp'}$ is orthogonal to $\vec{q_1q_2}$. Using these facts there can be solved an equation system.

What I try to understand:

However in my lecture I have read this:

$$\frac{1}{||v||}v \left( \frac{1}{||v||}v^{T}p\right) = \frac{1}{||v||^2}(vv^{T})p=\frac{vv^{T}}{v^{T}v}p$$

Question: What do these inner and outer products mean thus I will be able to understand that equation?

Note: I have seen similiar posts on that topic like this one. Unfotunately I have not found the formula above in none of them.

I appreciate any help!

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An inner product $\langle v,w\rangle$ can be define on $\mathbb{R}^n$ through vector multiplication: $\langle v,w\rangle = w^{\perp} v$. A general point on the line through two distinct points $p_1,p_2$ can be described by $p(t)=p_1+t(p_2-p1)=(1-t)p_1+tp_2$. If you want the orthogonal projection of $q$ onto such a line, then you search for the unique $p(t)$ such that $(q-p(t))\perp (p_2-p_1)$ because $p_2-p_1$ is the direction vector of the line. So, this amounts to finding $t$ such that

$$ (q-p_1-t(p_2-p_1))\perp (p_2-p_1) \\ \langle q-p_1,p_2-p_1\rangle - t\langle p_2-p_1,p_2-p_1\rangle = 0 \\ t = \frac{\langle q-p_1,p_2-p_1\rangle}{\langle p_2-p_1,p_2-p_1\rangle}. $$ So the orthogonal projection of $q$ onto the line is $p(t)$ where $t$ is given above.

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