Dot Product of Two vectors is commutative

Question

I was reading this book by Oleg D. Jefimenko on Electricity and Magnetism which had a chapter on Vector Analysis. In this section, he defined dot product as $\bar A \cdot \bar B$ as |$ {\bar A} $||$\bar B$| $cos \theta$ and also as $A_xB_x+A_yB_y+C_xC_y$ and he stated that dot product is commutative. And then he went used dot product to prove cosine law for triangles:

$$\bar C= \bar A+\bar B$$ $${\bar C}^2= (\bar A+\bar B)\cdot(\bar A+\bar B)$$ $${\bar C}^2= (\bar A \cdot \bar A)+(\bar A \cdot \bar B)+(\bar B \cdot \bar A)+(\bar B \cdot \bar B)$$ $${\bar C}^2= {\bar A}^2+{\bar B}^2+(\bar A \cdot \bar B)+(\bar B \cdot \bar A)$$

and then it went ahead and did something that left me confounded:

It said: $$ \bar A \cdot \bar B = |\bar A| |\bar B| cos \theta$$ $$ \bar B \cdot \bar A = |\bar B| |\bar A| cos (180^o -\theta)=-|\bar B| |\bar A| cos (\theta)$$ and then went ahead with the proof. Isn't this contradictory to the part where it said that dot product of commutative.

I know this law can be proved using alternate means by choosing to reorient the sides of the triangle and rewriting $\bar C$ as $\bar C = \bar A - \bar B$.

But I want to know how did it manage to disregard the commutative property of dot product.

By the way this is on page 26 of the second edition of the book.

Edit:

Answer 1

You are misunderstanding what he is saying. Note that he converts

• $\mathbf b \cdot \mathbf c = bc\cos(\mathbf b, \mathbf c)$
• $\mathbf c \cdot \mathbf b = cb\cos(\mathbf c, \mathbf b)$

And in the very next sentence, he clearly states: $$\angle (\mathbf b, \mathbf c) =\angle (\mathbf c, \mathbf b)$$ Which along with commutivity of the multiplication $bc = cb$ still leaves us with $$\mathbf b \cdot \mathbf c = \mathbf c \cdot \mathbf b$$

What he is saying is that neither of those angles is $\theta$. Instead they are both equal to $180^\circ - \theta$.

$\theta$ itself is the angle between $\mathbf c$ and $(-\mathbf b)$, the vector of the same length pointing in the opposite direction of $\mathbf b$. Remember that angles are calculated between vectors based at the same point. But in the picture, $\mathbf c$ has been slid so that its base is at the end of $\mathbf b$.