成分表示された平面ベクトルの減法

2 つのベクトルを，$\vec{a} = \left( \begin{array}{c} a_x\\ a_y\\ \end{array} \right) ，\vec{b} = \left( \begin{array}{c} b_x\\ b_y\\ \end{array} \right)$ とおくと，$−\vec{b} = \left( \begin{array}{c} −b_x\\ −b_y\\ \end{array} \right)$ となるから

\[ \vec{a} − \vec{b} = \left( \begin{array}{c} a_x\\ a_y\\ \end{array} \right) − \left( \begin{array}{c} b_x\\ b_y\\ \end{array} \right) = \left( \begin{array}{c} ax − bx\\ ay − by\\ \end{array} \right)\]

となる．

ベクトルの減法

無題

無題

平行四辺形$\text{ ABCD}$ の対角線の交点を$\text{ O}$ とし， $\overrightarrow{\text{OA}} = \vec{a} = \left( \begin{array}{c} −1\\ 1\\ \end{array} \right) ， \overrightarrow{\text{OB}} = \vec{b} = \left( \begin{array}{c} −2\\ −1\\ \end{array} \right)$ とする．次のベクトルを$\vec{a}，\vec{b}$ を用いて表し，また，成分表示せよ．

  1. $\overrightarrow{\text{DO}}$
  2. $\overrightarrow{\text{DA}}$
  3. $\overrightarrow{\text{AB}}$
  4. $\overrightarrow{\text{OC}}$
  5. $\overrightarrow{\text{BC}}$

  1. \[\overrightarrow{\text{DO}}=−\overrightarrow{\text{OD}}=−(−\vec{b})=\left(\begin{array}{c}\boldsymbol{−2}\\\boldsymbol{−1}\\\end{array}\right)\]
  2. \begin{align} \overrightarrow{\text{DA}}&=\overrightarrow{\text{OA}}−\overrightarrow{\text{OD}}=\vec{a}−(−\vec{b})\\ &=\left(\begin{array}{c}−1\\1\\\end{array}\right)−\left(\begin{array}{c}2\\1\\\end{array}\right)=\left(\begin{array}{c}\boldsymbol{−3}\\\boldsymbol{0}\\\end{array}\right) \end{align}
  3. \begin{align} \overrightarrow{\text{AB}}&=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}=\vec{b}−\vec{a}\\ &=\left(\begin{array}{c}-2\\-1\\\end{array}\right)-\left(\begin{array}{c}-1\\1\\\end{array}\right)=\left(\begin{array}{c}\boldsymbol{-1}\\\boldsymbol{-2}\\\end{array}\right) \end{align}
  4. \begin{align} \overrightarrow{\text{OC}}&=-\overrightarrow{\text{OA}}=-\vec{a}\\ &=-\left(\begin{array}{c}-1\\1\\\end{array}\right)=\left(\begin{array}{c}\boldsymbol{1}\\\boldsymbol{-1}\\\end{array}\right) \end{align}
  5. \begin{align} \overrightarrow{\text{BC}}&=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}=-\vec{a}-\vec{b}\\ &=-\left(\begin{array}{c}-1\\1\\\end{array}\right)-\left(\begin{array}{c}-2\\-1\\\end{array}\right)=\left(\begin{array}{c}\boldsymbol{3}\\\boldsymbol{0}\\\end{array}\right) \end{align}