벡터 합의 크기

좌표평면의 점 $\mathrm{A} \mathit{\left( 0 , 2 \right)}$와 원점 $\mathrm{O}$에 대하여 제$1$사분면의 점 $\mathrm{B}$를 잡을 때 삼각형 $\mathrm{AOB}$가 정삼각형이므로 $\displaystyle \mathrm{B} \mathit{\left( \sqrt{3} , 1 \right)}$

점 $\displaystyle \mathrm{C} \mathit{\left( - \sqrt{3} , 0 \right)}$에 대하여 ${\mathrm{\vec{BC}}}$$= \mathrm{\vec{OC}} \mathit{-} \mathrm{\vec{OB}}$$\displaystyle = \left( - \sqrt{3} , 0 \right) - \left( \sqrt{3} , 1 \right)$$\displaystyle = \left( - 2 \sqrt{3} , - 1 \right)$ $\therefore$$\displaystyle \mathrm{\vec{OA}} \mathit{+} \mathrm{\vec{BC}} \mathit{=} \left( 0 , 2 \right) + \left( - 2 \sqrt{3} , - 1 \right)$$\displaystyle = \left( - 2 \sqrt{3} , 1 \right)$ $\therefore$$\left| \mathrm{\vec{OA}} \mathit{+} \mathrm{\vec{BC}} \right|$$\displaystyle = \sqrt{\left( - 2 \sqrt{3} \right) ^{2} + 1 ^{2}}$$\displaystyle = \sqrt{13}$