## Calculus (3rd Edition)

$$v_1=Q-P=(4,4)-(2,4)=\langle 2, 0 \rangle, \quad \|v_1\|=\sqrt{4+0}=2.$$ $$v_2=Q-P=(1,3)-(-1,3)=\langle 2, 0 \rangle, \quad \|v_2\|=\sqrt{4+0}=2$$ $$v_3=Q-P=(2,4)-(-1,3)=\langle 3, 1 \rangle, \quad \|v_3\|=\sqrt{9+1}=\sqrt{10}$$ $$v_4=Q-P=(6,3)-(4,1)=\langle 2, 2 \rangle, \quad \|v_3\|=\sqrt{4+4}=\sqrt{8}$$ $v_1$ and $v_2$ are equivalent.
$$v_1=Q-P=(4,4)-(2,4)=\langle 2, 0 \rangle, \quad \|v_1\|=\sqrt{4+0}=2.$$ $$v_2=Q-P=(1,3)-(-1,3)=\langle 2, 0 \rangle, \quad \|v_2\|=\sqrt{4+0}=2$$ $$v_3=Q-P=(2,4)-(-1,3)=\langle 3, 1 \rangle, \quad \|v_3\|=\sqrt{9+1}=\sqrt{10}$$ $$v_4=Q-P=(6,3)-(4,1)=\langle 2, 2 \rangle, \quad \|v_3\|=\sqrt{4+4}=\sqrt{8}$$ $v_1$ and $v_2$ are equivalent.