Calculating Sum of 2 Vectors using Laws of Cosine

Question

I found this to calculate the sum of 2 vectors with a specific angle $v$:

It's the law of cosine: $$a^2 + b^2 - 2ab\cos(v)$$

Sources are split on this, however ...

One source says the one above is the way to go, but others say this one is:

$$a^2 + b^2 +2ab\cos(v) $$

(the same but with + and + instead of + and -)

Could someone please shed a light on this?

Answer 1

It depends if the angle $v$ is between the two vectors taken ''tail to tail'' or ''head to tail''. This two angles are supplementary. You can see this answer: Proof of vector addition formula

Answer 2

There is an ambiguity here which I think comes from the way we define the "angle" that is used in the formula.

Consider this visualization of vector addition, copied from https://en.wikipedia.org/wiki/File:Vector_Addition.svg:

To find the length of $\mathbf a + \mathbf b$, we can apply the Law of Cosines to one of the two triangles with sides $\mathbf a$, $\mathbf b$, and $\mathbf a + \mathbf b$. Either triangle is OK since they are congruent. The Law of Cosines tells us that $$ c^2 = a^2 + b^2 - 2ab \cos \phi $$ where $a = \|\mathbf a\|$, $b = \|\mathbf b\|$, $c = \|\mathbf a + \mathbf b\|$, and $\phi$ is the angle of the sides $\mathbf a$ and $\mathbf b$ of the triangle. Depending on which of the two triangles you choose, this is either the angle where the head of a copy of $\mathbf a$ meets the tail of a copy of $\mathbf b$ or the angle where the head of a copy of $\mathbf b$ meets the tail of a copy of $\mathbf a$. In other words, to use this formula we should set $\phi$ to one of the "head-to-tail" angles in the figure.

But when we measure the angle between two vectors, it is more usual to measure the "tail-to-tail" angle, that is, the angle in the figure where the tail of a copy of $\mathbf a$ and the tail of a copy of $\mathbf b$ coincide (which is also the point where the tail of $\mathbf a + \mathbf b$ is drawn in the figure).

Let's write $\alpha$ to denote the "tail-to-tail" angle between the vectors. Since the copies of $\mathbf a$ and $\mathbf b$ in the figure above form a parallelogram, we know that $$ \alpha = \pi - \phi $$ (measured in radians) and therefore $$ \cos(\alpha) = -cos(\phi). $$ Therefore the Law of Cosines actually tells us that $$ c^2 = a^2 + b^2 + 2ab \cos \alpha. $$

In other words, whether you have "$+\cos\theta$" or "$-\cos\theta$" is entirely due to whether you define $\theta$ the way $\phi$ is defined above or the way $\alpha$ is defined above.

The answer Proof of vector addition formula comes to this conclusion as well, but the figure that supports that answer seems to have been lost.