Áp dụng quy tắc hình bình hành, ta có: $\vec{p}$ = $\vec{AB}$ + $\vec{AD}$ = $\vec{AC}$
$\Rightarrow$ |$\vec{p}$| = |$\vec{AC}$| = $\sqrt{AB^{2} + AD^{2} + 2.AB.AD. cos\widehat{BAD}}$ =
$\sqrt{a^{2} + a^{2} + 2. a. a.cos60^{\circ}} = \sqrt{3}a$
Ta có: $\vec{u}$ = $\vec{AB}$ - $\vec{AD}$ = $\vec{DB}$
$\Rightarrow$ |$\vec{u}$| = |$\vec{DB}$| = BD = a
Ta có: $\vec{v}$ = 2$\vec{AB}$ - $\vec{AC}$ = $\vec{AB}$ - $\vec{AC}$ + $\vec{AB}$ = $\vec{CB }$ + $\vec{AB}$ = $\vec{DA}$ + $\vec{AB}$ = $\vec{DB}$
$\Rightarrow$ |$\vec{v}$| = |$\vec{DB}$| = BD = a.