Vector Calculus (8) - Scalar Product

19. Scalar Product(Dot Product)

 

$ \vec{A}\cdot \vec{B}=\sum_{i}^{}A_{i}B_{i}=AB\cos (\vec{A},\vec{B}) $

$ |\vec{A}|=\sqrt{A_{1}^{2}+A_{2}^{2}+A_{3}^{2}}\equiv A $: Magnitude(Length) of $ \vec{A} $

$ \frac{A_{i}}{A}=\Lambda_{i}^{A}, \frac{B_{i}}{B}=\Lambda_{i}^{B} $: Direction Cosines of $ \vec{A} $ and $ \vec{B} $

$ \cos \Theta =\cos (\vec{A},\vec{B})=\cos \alpha \cos \alpha'+\cos \beta+\cos \beta '+\cos \gamma \cos \gamma ' $

 

$ \frac{\vec{A}\cdot \vec{B}}{AB}=\sum_{i}^{}\frac{A_{i}}{A}\frac{B_{i}}{B}=\sum_{i}^{}\Lambda_{i}^{A}\Lambda_{i}^{B}=\cos (\vec{A},\vec{B}) $

$ \Rightarrow  \vec{A}\cdot \vec{B}=AB\cos (\vec{A},\vec{B}) $

 

$ \vec{A}\cdot \vec{B}=\sum_{i}^{}A_{i}B_{i}=\sum_{i}^{}B_{i}A_{i}=\vec{B}\cdot \vec{A} $: Commutative Law

$ \vec{A}\cdot (\vec{B}+\vec{C})=\sum_{i}^{}A_{i}(B+C)_{i}=\sum_{i}^{}A_{i}(B_{i}+C_{i})=\sum_{i}^{}A_{i}B_{i}+A_{i}C_{i}=\vec{A}\cdot \vec{B}+\vec{A}\cdot \vec{C} $: Distributive Law

 

 

20. Result of Scalar Product is Scalar

 

$ A_{i}'=\sum_{j}^{}\lambda _{ij}A_{j}, B_{i}'=\sum_{k}^{}\lambda _{ik}B_{k} $

$ \vec{A}'\cdot \vec{B}'=\sum_{i}^{}A_{i}'B_{i}'=\sum_{i}^{}(\sum_{j}^{}\lambda _{ij}A_{j})(\sum_{k}^{}\lambda _{ik}B_{k})=\sum_{j,k}^{}(\sum_{i}^{}\lambda _{ij}\lambda _{ik})A_{j}B_{k} $

  $ =\sum_{j}^{}(\sum_{k}^{}\delta _{jk}A_{j}B_{k})=\sum_{j}^{}A_{j}B_{j}=\vec{A}\cdot \vec{B} $

→ Scalar (The value of the product is unaltered by the coordinate transformation)