I'm trying to solve these questions:

Given $v = [3, -6, 2]$ and $w = [-1, 6, 5]$, find;

  1. $v \downarrow w$
  2. $w \downarrow v$
  3. What does the magnitude of $w \downarrow v$ depend on?
  4. What does the direction of $w \downarrow v$ depend on?

I already figured out the answer to 1 and 2, since they are pretty straightforward. But I don't really know the answer to 3 and 4.

  • $\begingroup$ They both depend on the angle between the two vectors (in the plane generated by those two vectors). the magnitude depends on how far the angle is from being a right angle. And the direction depends on whether the angle is less than 90 degrees or larger. Do you know what a scalar product is? $\endgroup$ – xavierm02 Jan 21 '14 at 17:27
  • $\begingroup$ I don't think so. I know what a dot product and cross product is, at least. $\endgroup$ – Threethumb Jan 21 '14 at 17:30
  • $\begingroup$ A scalar is a dot product. $\endgroup$ – xavierm02 Jan 21 '14 at 17:31
  • $\begingroup$ Ah, okay. So how does scalar/dot products relate to the answer here? $\endgroup$ – Threethumb Jan 21 '14 at 17:35

When one speaks of projecting $\vec a$ onto $\vec b$ (I will write $p_{\vec b}(\vec a)$), I think it is understood that you really mean "the projection of $\vec a$ in the direction of $\vec b$", that, is $\vec b$ is only used to get a direction, and its (nonzero) magnitude is irrelevant.

The direction of $p_{\vec b}(\vec a)$ depends only the direction of $\vec b$ (in fact, it is the direction of $\vec b$).

The magnitude of $p_{\vec b}(\vec a)$ depends on the relative direction of $\vec a$ with respect to $\vec b$ (said differently, the angle between $\vec a$ and $\vec b$) as well as the magnitude of $\vec a$.

  • $\begingroup$ I read up on projections, and understand your answer now. I can see why it is the right answer too, now. $\endgroup$ – Threethumb Jan 23 '14 at 11:27
  • $\begingroup$ Glad to hear that, thanks. You can see the correctness of the terminology--it's the "shadow" of the projected vector on the space generated by the "target". Good for you! $\endgroup$ – MPW Jan 25 '14 at 7:07

Let $p_\vec{y}$ be the projection on the line generated by $\vec{y}$, that is $$p_\vec{y}:\begin{array}{l}\Bbb R^3 \to \Bbb R^3\\\vec{x}\mapsto \cfrac{\langle \vec{x} \mid \vec{y}\rangle}{\|\vec{x}\|\|\vec{y}\|}\vec{y} \end{array}$$

$\left\|p_\vec{y}(\vec{x})\right\|=\left|\cfrac{\langle \vec{x} \mid \vec{y}\rangle}{\|\vec{x}\|\|\vec{y}\|}\|\vec{y}\|\right|=\cfrac{\left|\cos\left( \vec{x} , \vec{y}\right)\|\vec{x}\|\|\vec{y}\|\right|}{\|\vec{x}\|\|\vec{y}\|}\|\vec{y}\|=\left|\cos\left(\vec{x},\vec{y}\right)\right|\|\vec y\|$ so it depends on the cosinus of the angle between $\vec x$ and $\vec y$.

Its direction depends on the sign of $\langle \vec{x} \mid \vec{y}\rangle$. If it is positive, it has the same direction as $y$ and otherwise, it has an opposite direction.


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