As you can see below, $z$ is the projection of $x$ onto $y$...

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I am trying to derive the orthogonal projection formula based on things I already know.

Calculating $cos(\theta)$ is trivial...

$$cos(\theta) = \frac{\left \| z \right \|}{\left \| x \right \|}$$ $$\left \| z \right \| = \left \| x \right \|cos(\theta )$$

I also know that the dot product can be re-written as...

$$cos(\Theta ) = \frac{xy}{\left \| x \right \| \left \| y \right \|}$$

So by substitution we get...

$$\left \| z \right \| = \frac{xy}{\left \| y \right \|}$$

We can define the direction of $y$ as...

$$u = \frac{y}{\left \| y \right \|}$$

So we now have...

$$\left \| z \right \| = u\cdot x$$

Now since $z$ and $y$ point in the same direction, $u$ can be written as...

$$u = \frac{z}{\left \| z \right \|}$$ $$z = \left \| z \right \|\cdot u$$

Therefore, we have...

$$z = (u \cdot x) u$$

What I find strange is the following two equations...

$$\left \| z \right \| = u\cdot x$$ $$z = (u \cdot x) u$$

What this is telling me is that, multiply $x$ by the y-direction, $u$, and you get the length of $z$, in other words, $\left \| z \right \|$. Multiply $x$ by the y-direction, $u$, again, and you get the vector $z$ with direction.

I can't seem to wrap my head around why this is true. Is there a more intuitive explanation as to why multiplying $x$ by y-direction goes from getting the length of $z$ to getting the $z$ vector itself?

  • $\begingroup$ There are two different types of multiplication in $(u\cdot x)u$, but you appear to be interpreting both the same way. $\endgroup$ – amd Apr 16 at 5:50
  • $\begingroup$ @amd two different ones in what sense? $\endgroup$ – Bolboa Apr 16 at 12:52
  • $\begingroup$ $u\cdot x$ takes two vectors and produces a scalar $a$. You then multiply the vector $u$ by this scalar to get another vector. These two operations are rather different. $\endgroup$ – amd Apr 16 at 16:58

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