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$\cos \theta = \frac{av}{|a||v|}$ (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)


$a_v v = |a_v||v|\cos0=|a_v||v|$ (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.

  • $\begingroup$ $a_v=\lVert a\rVert\cos{\theta}$ in the direction of $v$ which is given by $\frac{1}{\lVert v\rVert}v$. $\endgroup$
    – John Douma
    May 31 at 15:26
  • 1
    $\begingroup$ 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 $\endgroup$ May 31 at 15:37

2 Answers 2


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 $|a|\neq 0$

$$\frac{a\cdot v}{|a||v|}= \frac{|a_v|}{|a|} \implies a\cdot v=|a_v||v|=|a||v|\cos \theta$$


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