# where I'm wrong with direct cosine vs dot product?

$$\cos \theta = \frac{av}{|a||v|}$$ (1) (from dot product)

$$\cos \theta = \frac{|a_v|}{|a|}$$ (2) (from cosine definition)

combining (1) and (2):

$$\frac{av}{|a||v|} = \frac{|a_v|}{|a|}$$, so $$av = |a_v||v|$$ (3)

but

$$a_v v = |a_v||v|\cos0=|a_v||v|$$ (4) (by dot product)

now if combine (3) and (4), I have:

$$av=a_vv$$ and $$a = a_v$$ which is false by assumption.

What's wrong? There must be some obvious explanation, but I checked it multiple times and can't see it.

• $a_v=\lVert a\rVert\cos{\theta}$ in the direction of $v$ which is given by $\frac{1}{\lVert v\rVert}v$. May 31 at 15:26
• It would be better (in my opinion), that for example, instead of $av,$ you wrote $a\ .\ v,\$ to clarify that you're using the dot product. Don't sacrifice clarity for brevity... By the way, as a side note, it is exactly the observation you make about $a\ .\ v = a_v\ .\ v\$ which is why the equation of a plane can be written as $r\ .\ n = k,\$ where $n$ is a normal vector to the plane, and $k$ is a constant. Then, every triplet $r = \begin{bmatrix} x \\ y\\ z \end{bmatrix}$ that satisfies the above equation are precisely the points on the plane May 31 at 15:37

You correctly write $$av=a_vv$$. But you forget that these are dot products and you cannot just divide away the $$v$$ from each side. $$av=a_vv$$ does not imply that $$a=a_v$$. It does imply that $$(a-a_v)v=0$$, so the vectors $$a-a_v$$ and $$v$$ are orthogonal, which is consistent with your diagram.
Note that $$\cos \theta = \frac{|a_v|}{|a|}$$ is true only for $$\cos \theta \ge 0$$, and in this case we have for $$|a|\neq 0$$
$$\frac{a\cdot v}{|a||v|}= \frac{|a_v|}{|a|} \implies a\cdot v=|a_v||v|=|a||v|\cos \theta$$