## Calculus: Early Transcendentals 8th Edition

Since, $u.v=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}$ and $v.u=v_{1}u_{1}+v_{2}u_{2}+v_{2}u_{2}$ Consider $u=\lt 1,0,0\gt$ and $v=\lt 0,1,0\gt$ $u.v=1*0+0*1=0+0*0=0$ and $v.u=0*1+1*0+0*0=0$ Thus, $u.v= v.u$ Hence, the statement is true.