## Thomas' Calculus 13th Edition

Counterexample: Let ${\bf u}=\langle 1,1\rangle$ ${\bf v_{1}}=\langle 1,0\rangle,\quad {\bf v_{2}}=\langle 0,1\rangle$ ${\bf u}\cdot{\bf v_{1}}=1+0=1,$ ${\bf u}\cdot{\bf v_{2}}=0+1=1.$ ${\bf u}\cdot{\bf v_{1}}={\bf u}\cdot{\bf v_{2}}$, but ${\bf v_{1}}\neq{\bf v_{2}}$