Question #206258

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


1
Expert's answer
2021-06-15T10:37:17-0400

Assume two vectors u=xi^+yj^+zk^,v=ai^+bj^+ck^\vec u=x \hat i+y\hat j+z\hat k, \vec v=a \hat i+b\hat j+c\hat k .


Now, u.v=(xi^+yj^+zk^).(ai^+bj^+ck^)\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=xa+yb+zc ...(i)


And v.w=(ai^+bj^+ck^).(xi^+yj^+zk^)\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=ax+by+cz ...(ii)


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


Hence, commutative.


Also, assume another vector w=pi^+qj^+rk^\vec w=p \hat i+q\hat j+r\hat k

(u.v).w=[(xi^+yj^+zk^).(ai^+bj^+ck^)].(pi^+qj^+rk^)=[xa+yb+zc].(pi^+qj^+rk^)(\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,

u.(v.w)=(xi^+yj^+zk^).[(ai^+bj^+ck^).(pi^+qj^+rk^)]=(xi^+yj^+zk^).(ap+bq+cr)\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.


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