Theoretical Mechanics IPSP

Jürgen Vollmer, Universität Leipzig

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book:chap2:2.7_the_inner_product

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book:chap2:2.7_the_inner_product [2024/12/03 23:06] jvbook:chap2:2.7_the_inner_product [2024/12/03 23:28] (current) jv
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     c^2 = a^2 + b^2 - 2\,a\,b\,\cos\theta     c^2 = a^2 + b^2 - 2\,a\,b\,\cos\theta
 \end{align*} \end{align*}
-Let now $a$, $b$, and $c$ be the length of the vectors $\mathbf a$, $\mathbf b$ and $\mathbf c = \mathbf - \mathbf b$,+Let now $a$, $b$, and $c$ be the length of the vectors $\mathbf a$, $\mathbf b$ and $\mathbf c = \mathbf - \mathbf a$,
 as shown in [[#fig_scalarProductCosSetup |Figure 2.13]]. as shown in [[#fig_scalarProductCosSetup |Figure 2.13]].
 Then we have Then we have
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     a^2 + b^2 - 2\,a\,b\,\cos\theta     a^2 + b^2 - 2\,a\,b\,\cos\theta
     &= c^2 = \mathbf c \cdot \mathbf c     &= c^2 = \mathbf c \cdot \mathbf c
 +      = (\mathbf b - \mathbf a) \cdot  (\mathbf b - \mathbf a)
       = (\mathbf a - \mathbf b) \cdot  (\mathbf a - \mathbf b)       = (\mathbf a - \mathbf b) \cdot  (\mathbf a - \mathbf b)
     \\     \\
     &= \mathbf a \cdot \mathbf a - 2\, \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf b     &= \mathbf a \cdot \mathbf a - 2\, \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf b
       = a^2 + b^2 - 2\, \mathbf a \cdot \mathbf b       = a^2 + b^2 - 2\, \mathbf a \cdot \mathbf b
-    \\+    \\[2mm]
     \Rightarrow\quad     \Rightarrow\quad
-    \mathbf a \cdot \mathbf b = |\mathbf a| \, |\mathbf b| \, \cos\theta+    \mathbf a \cdot \mathbf b &= |\mathbf a| \, |\mathbf b| \, \cos\theta
     \\     \\
 \end{align*} \end{align*}
book/chap2/2.7_the_inner_product.1733263614.txt.gz · Last modified: 2024/12/03 23:06 by jv