Angle between these two vectors

Question

A mapping tool moves from point $a$ {199,176} to point $b$ {199,164} then moves to point $c$ {198, 185} the movement coordinates in the image shows this.

the question is what is the angle that creates between these vectors at {199, 164}

Answer 1

There are multiple solutions for a particular cosine value. For example, for a negative cos value, there are two solutions in the domain $\left \{ \pi /2 \leq x \leq 3\pi/2 \right \}$

If you use the rule:

$a\cdot b = \left | a \right |\left | b \right |\cos (\theta )$

With the vectors represented as:

$$a = 199176\textbf{i} -199164\textbf{j} $$ $$b = 199164\textbf{i} - 198185\textbf{j} $$

This will give you an angle of $0.139441°$

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For your updated question I would use $\vec{ab}$ and $\vec{cb}$. Then you can plug this into the dot product rule as with the original solution. From what I can grasp from your question these are the vectors you would substitute:

$$\vec{ab} = -12\textbf{j}$$ $$\vec{cb} = 1\textbf{i} -21\textbf{j}$$

For the angle between these two I got approx. $2.72°$

Hopefully this helps :)

Answer 2

drawing of the solution

Definition

The angle between two vectors, deferred by a single point, called the shortest angle at which you have to turn around one of the vectors to the position of co-directional with another vector.

I arrived at the answer inspired by this answer provided by DonAntonio. In the formula, the absolute value of the cosine is used.

$\vec ab = (0, -12)\\\vec bc = (-1, 21)\\dot product = \vec ab.\vec bc = -252$

magnitude of vectors = $\left\lvert ab\right\rvert.\lvert bc \rvert = 252.28\\$

cosine of the angle between two vectors $\left(\frac{\vec ab . \vec bc }{\lvert ab \rvert . \lvert bc \rvert}\right) = 0.99889$

$cos^\text{-1} of 0.99889*(180/\pi) = 2.69969951^\circ$ (degrees)

since we have added the vectors end of the first vector to the beginning of second vector so to calculate the angle formed between these added vectors we need to subtract above value from 180 degrees.