The angle #\varphi# is the short angle made by #\vec{u}# and #\vec{v}#; more precisely: the two representatives made by these vectors with the same starting point. In particular, the following applies: #0^\circ\leq\varphi\leq180^\circ#. The oriented angle, defined in the plane by the "counterclockwise"orientation, is not determined in the space: after all, we cannot properly determine whether we are viewing a plane from above or from below; we can, therefore, not agree on the orientation.

Example: if the vectors #\vec{u}# and #\vec{v}# both have a length of#4#, and the angle between the two vectors is equal to #60^{\circ}#, then
\[\dotprod{\vec{u}}{\vec{v}}= 4\cdot 4\cdot \cos\left( 60^{\circ}\right) = 4\cdot 4\cdot \frac{1}{2} = 8
\]
The dot product is #-8# for an angle of #120^{\circ}#. The dot product can, in fact, be negative.

The literature also contains other notations such as #\left(\vec{u},\vec{v}\right)# or #\langle\vec{u},\vec{v}\rangle#.

The dot product is also referred to as dot product or scalar product.

First, suppose that #A# is on #\ell#. In that case #A=P#, and #\triangle ABQ# is a right-angled triangle. Let #\varphi# be the angle between the vectors #\vec{v}=\vec{AB}# and #\vec{u}#. In that case \[\vec{PQ}=\varepsilon \cdot \frac{\parallel \vec{PQ}\parallel}{\parallel \vec{u}\parallel}\vec{u}\] where #\varepsilon=1# if #\vec{PQ}# and #\vec{u}# point in the same direction, and #-1# if they point in a different one. Inspection of the triangle #\triangle ABQ# shows that \[\cos(\varphi)=\varepsilon\cdot\frac{\parallel \vec{PQ}\parallel}{\parallel \vec{v}\parallel}\tiny.\] If #\vec{v}=\vec{0}#, then both sides of the equality we are trying to prove, are the zero vector. In that case, we can assume that #\vec{v}# is not the zero vector, so #\parallel\vec{v}\parallel\ne0#. From here, we conclude that

\[\begin{array}{rcll}\vec{PQ}&=&\varepsilon \cdot \frac{\parallel \vec{PQ}\parallel}{\parallel \vec{u}\parallel}\cdot\vec{u}&\phantom{xxx}\color{blue}{\text{the formula above for }\vec{PQ}}\\ &=&\varepsilon\cdot\frac{\parallel \vec{PQ}\parallel}{\parallel \vec{v}\parallel}\cdot \frac{\parallel \vec{v}\parallel}{\parallel\vec{u}\parallel}\cdot\vec{u}&\phantom{xxx}\color{blue}{\text{scalar rewritten }}\\&=&\cos(\varphi)\cdot \frac{\parallel \vec{v}\parallel}{\parallel \vec{u}\parallel}\cdot\vec{u}&\phantom{xxx}\color{blue}{\text{formula above for }\cos(\varphi)}\\ &=& \dfrac{\vec{u}\cdot\vec{v}}{\parallel\vec{u}\parallel\cdot\parallel\vec{v}\parallel} \cdot\frac{\parallel \vec{v}\parallel}{\parallel \vec{u}\parallel}\cdot\vec{u}&\phantom{xxx}\color{blue}{\text{the dot product in terms of lengths}}\\&=& \dfrac{\vec{u}\cdot\vec{v}}{\parallel\vec{u}\parallel^2} \cdot\vec{u}&\phantom{xxx}\color{blue}{\text{simplified}}\end{array}\] This is the formula if #A# is on #\ell#. If this is not the case, then we can compare the initial situation with the situation in which the vector #\vec{v}# has shifted over #\vec{PA}#. The projection of the starting point of the new representative of #\vec{v}# must then be #P#, and the projection of the end point of the new representative of #\vec{v}# must then be #Q#, because #\vec{v}#, and therefore #B#, is displaced in a direction perpendicular to #\ell#. The left and right sides of the formula we are trying to prove in the new situation, do not differ from the old situation. But the equality is proven in the new situation, and must therefore also apply in the old situation. This is the completed proof of the formula.

The independence of the vector #\vec{PQ}# from the placement of #\vec{v}# will be used later.