Answer to Question #206258 in Analytic Geometry for Casey

Question

Prove that the dot product between two vectors is commutative and not associative

Answer 1

Assume two vectors $\vec u=x \hat i+y\hat j+z\hat k, \vec v=a \hat i+b\hat j+c\hat k$ .

Now, $\vec u.\vec v=(x \hat i+y\hat j+z\hat k).(a \hat i+b\hat j+c\hat k)$$=xa+yb+zc$ ...(i)

And $\vec v.\vec w=(a \hat i+b\hat j+c\hat k).(x \hat i+y\hat j+z\hat k)$$=ax+by+cz$ ...(ii)

From (i) and (ii), $\vec u.\vec v=\vec v.\vec u$

Hence, commutative.

Also, assume another vector $\vec w=p \hat i+q\hat j+r\hat k$

$(\vec u.\vec v).\vec w=[(x \hat i+y\hat j+z\hat k).(a \hat i+b\hat j+c\hat k)].(p \hat i+q\hat j+r\hat k) \\=[xa+yb+zc].(p \hat i+q\hat j+r\hat k)$

which is not defined as dot product of a scalar and vector quantity is not defined.

Now,

$\vec u.(\vec v.\vec w)=(x \hat i+y\hat j+z\hat k).[(a \hat i+b\hat j+c\hat k).(p \hat i+q\hat j+r\hat k)] \\=(x \hat i+y\hat j+z\hat k).(ap+bq+cr)$

which is not defined as dot product of a scalar and vector quantity is not defined.

Hence, not associative.