Dot Product as a Vector

The dot product is a scalar while the cross product is a vector. Normally with a dot product, you do

$<x_1,y_1>\cdot<x_2,y_2>=x_1\cdot x_2+y_1\cdot y_2$

It can be thought of as how parallel two vectors are. The other equation for the dot product is

$|v_1||v_2|cos(\theta)$

If the vectors are parallel, the dot product is maximum and equals the lengths of the vectors multiplied by each other. So, an equation for two vector’s parallelism could be

$\frac{v_1\cdot v_2}{|v_1||v_2|}=cos(\theta)$
I was wondering what would happen if the dot product was a vector. I tried doing

$<x_1,y_1>\cdot<x_2,y_2>=<x_1\cdot x_2,y_1\cdot y_2>$

I did a test with one vector <5,3> and the other <2,3>. The slopes were .6 and 1.5. The dot product vector was <10,9>. The angle between that vector and the x axis was
$arctan(\frac{9}{10})= 42 \textdegree$
The general equation for that angle is

$arctan(m_1\cdot m_2)$

where m is the slope of a vector.

I also tried using the average of the slopes to get a vector that was exactly in between, as judged by the angle. In this case the vector could be <2,2.1>. The angle is then $46.4 \textdegree$ . The general equation is

$arctan(\frac{1}{2}(m_1+m_2))$

However, the slope is not directly proportional to the angle. It’s related by arctan which is not linear after a while. An increase in slope from 5 to 10 doesn’t change the angle much but an increase from 1 to 2 does change the angle much.

The correct way of calculating the slope halfway between the two lines is

$\frac{1}{2}(arctan(\frac{3}{5})+arctan(\frac{3}{2}))= 43.6 \textdegree$

Was it a coincidence that the first two ways were close to the correct answer?

Figure 1.png

With a set slope for the first vector of 0.6, increasing the slope of the second vector increases the angle that is halfway between them. However, the angle doesn’t change much when the slope of the second vector gets high as mentioned earlier. The average slopes method and the dot product method are both close to the correct way at the intial $m_2$ I had picked of 1.5. It was just a coincidence that the angles were so close together.