Processing math: 100%

方向余弦的平方和为1的推导和比较

非零向量$r$与三条坐标轴的夹角$\alpha ,\beta ,\gamma$称为向量$r$的方向角.设$\overrightarrow {OM} = r = \left( {x,y,z} \right)$，

$$\begin{gathered} \cos \alpha = \frac{x}{{\left| {OM} \right|}} \hfill \\ \cos \beta = \frac{y}{{\left| {OM} \right|}} \hfill \\ \cos \gamma = \frac{z}{{\left| {OM} \right|}} \hfill \\ \end{gathered}$$

称为向量$r$的方向余弦.那么

$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$$

有2种方法得到上述结果.

方法1

因为

$$\begin{gathered} \left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right) \hfill \\ = \left( {\frac{x}{{\left| r \right|}},\frac{y}{{\left| r \right|}},\frac{z}{{\left| r \right|}}} \right) \hfill \\ = \frac{1}{{\left| r \right|}}\left( {x,y,z} \right) \hfill \\ = \frac{r}{{\left| r \right|}} = {e_r} \hfill \\ \end{gathered}$$

所以以向量$r$的方向余弦为坐标的向量是与$r$同方向的单位向量.由单位向量的模为1，得

$$\eqalign{ & \sqrt {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma } = 1 \cr & {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \cr}$$

方法2

$$\begin{gathered} {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma \hfill \\ = {\left( {\frac{x}{{\left| r \right|}}} \right)^2} + {\left( {\frac{y}{{\left| r \right|}}} \right)^2} + {\left( {\frac{z}{{\left| r \right|}}} \right)^2} \hfill \\ = \frac{{{x^2} + {y^2} + {z^2}}}{{{{\left( {\left| r \right|} \right)}^2}}} \hfill \\ = \frac{{{x^2} + {y^2} + {z^2}}}{{{x^2} + {y^2} + {z^2}}} \hfill \\ = 1 \hfill \\ \end{gathered}$$

我们给出方向余弦的定义的最初目的，是为了求与任意非零向量 同向的单位向量，因此，为了方便记忆，第一种方法较好.