合成ベクトルの内積 ~ 数学について考えてみる

2022年7月27日

合成ベクトルの内積

PLUMBAGO

| ⃗ a + ⃗ b |^{2} = | ⃗ a |^{2} + 2 ⃗ a \cdot ⃗ b + | ⃗ b |^{2}

$|\vec{a}+\vec{b}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2$

となることを確かめてみます。

⃗ a = (a_{1}, a_{2}), ⃗ b = (b_{1}, b_{2})

$\vec{a}=(a_1,a_2),\vec{b}=(b_1,b_2)$ とすると

\begin{matrix} ⃗ a + ⃗ b & = (a_{1}, a_{2}) + (b_{1}, b_{2}) = (a_{1} + b_{1}, a_{2} + b_{2}) \end{matrix}

$\begin{align*}\vec{a}+\vec{b}&=(a_1,a_2)+(b_1,b_2)\\[0.5em]&=(a_1+b_1,a_2+b_2)\end{align*}$

ベクトルの内積の性質より

| ⃗ a + ⃗ b |^{2} = (⃗ a + ⃗ b) \cdot (⃗ a + ⃗ b)

$|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})$

だから、成分から内積を求める公式

\begin{matrix} ⃗ p = (p_{1}, p_{2}), & ⃗ q = (q_{1}, q_{2}) の と き ⃗ p \cdot ⃗ q & = p_{1} q_{1} + p_{2} q_{2} \end{matrix}

$\begin{align*}\vec{p}=(p_1,p_2),&\vec{q}=(q_1,q_2)のとき\\ \vec{p}\cdot\vec{q}&=p_1q_1+p_2q_2\end{align*}$

をもちいると

\begin{matrix} | ⃗ a + ⃗ b |^{2} & = (a_{1} + b_{1})^{2} + (a_{2} + b_{2})^{2} = ({a_{1}}^{2} + 2 a_{1} b_{1} + {b_{1}}^{2}) + ({a_{2}}^{2} + 2 a_{2} b_{2} + {b_{2}}^{2}) = ({a_{1}}^{2} + {a_{2}}^{2}) + 2 (a_{1} b_{1} + a_{2} b_{2}) + ({b_{1}}^{2} + {b_{2}}^{2}) = | ⃗ a |^{2} + 2 ⃗ a \cdot ⃗ b + | ⃗ b |^{2} \end{matrix}

$\begin{align*}|\vec{a}+\vec{b}|^2&=(a_1+b_1)^2+(a_2+b_2)^2\\[0.5em]&=\left({a_1}^2+2a_1b_1+{b_1}^2\right)+\left({a_2}^2+2a_2b_2+{b_2}^2\right)\\[0.5em]&=\left({a_1}^2+{a_2}^2\right)+2\left(a_1b_1+a_2b_2\right)+\left({b_1}^2+{b_2}^2\right)\\[0.5em]&=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2\end{align*}$

　また、

⃗ a = (a_{1}, a_{2}), ⃗ b = (b_{1}, b_{2}), ⃗ c = (c_{1}, c_{2}), ⃗ d = (d_{1}, d_{2})

$\vec{a}=(a_1,a_2),\vec{b}=(b_1,b_2),\vec{c}=(c_1,c_2),\vec{d}=(d_1,d_2)$ の4ベクトルをもちいて

(⃗ a + ⃗ b) \cdot (⃗ c + ⃗ d)

$(\vec{a}+\vec{b})\cdot(\vec{c}+\vec{d})$ がどうなるかを調べてみると

\begin{matrix} ⃗ a + ⃗ b & = (a_{1} + b_{1}, a_{2} + b_{2}) ⃗ c + ⃗ d & = (c_{1} + d_{1}, c_{2} + d_{2}) \end{matrix}

$\begin{align*}\vec{a}+\vec{b}&=(a_1+b_1,a_2+b_2)\\[1em]\vec{c}+\vec{d}&=(c_1+d_1,c_2+d_2)\end{align*}$

であるから

\begin{matrix} (⃗ a + ⃗ b) \cdot (⃗ c + ⃗ d) = & (a_{1} + b_{1}) (c_{1} + d_{1}) + (a_{2} + b_{2}) (c_{2} + d_{2}) = & (a_{1} c_{1} + a_{1} d_{1} + b_{1} c_{1} + b_{1} d_{1}) + (a_{2} c_{2} + a_{2} d_{2} + b_{2} c_{2} + b_{2} d_{2}) = & (a_{1} c_{1} + a_{2} c_{2}) + (a_{1} d_{1} + a_{2} d_{2}) + (b_{1} c_{1} + b_{2} c_{2}) + (b_{1} d_{1} + b_{2} d_{2}) = & ⃗ a \cdot ⃗ c + ⃗ a \cdot ⃗ d + ⃗ b \cdot ⃗ c + ⃗ b \cdot ⃗ d \end{matrix}

$\begin{align*}&(\vec{a}+\vec{b})\cdot(\vec{c}+\vec{d})\\[0.5em]=&(a_1+b_1)(c_1+d_1)+(a_2+b_2)(c_2+d_2)\\[0.5em]=&(a_1c_1+a_1d_1+b_1c_1+b_1d_1)\\ &\qquad+(a_2c_2+a_2d_2+b_2c_2+b_2d_2)\\[0.5em]=&(a_1c_1+a_2c_2)+(a_1d_1+a_2d_2)\\ &\qquad+(b_1c_1+b_2c_2)+(b_1d_1+b_2d_2)\\[0.5em]=&\vec{a}\cdot\vec{c}+\vec{a}\cdot\vec{d}+\vec{b}\cdot\vec{c}+\vec{b}\cdot\vec{d}\end{align*}$

このように2つの二項式の積