Talk:PlanetPhysics/Dot Product Example

Original TeX Content from PlanetPhysics Archive
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\begin{document}

Examples involving the \htmladdnormallink{dot product}{http://planetphysics.us/encyclopedia/DotProduct.html}:

(1) Calculate $ \mathbf{A} \cdot \mathbf{B} $ with

$ \mathbf{A} = 5 \mathbf{\hat{i}} + 2 \mathbf{\hat{j}} - \mathbf{\hat{k}}$ \\ $ \mathbf{B} = 3 \mathbf{\hat{i}} - 3 \mathbf{\hat{j}} + 3\mathbf{\hat{k}}$ \\

answer:

$ \mathbf{A} \cdot \mathbf{B} = (5)(3) + (2)(-3) + (-1)(+3) $ \\ $ \mathbf{A} \cdot \mathbf{B} = 15 - 6 - 3 = 6$ \\

(2) Find the angle between the above \htmladdnormallink{vectors}{http://planetphysics.us/encyclopedia/Vectors.html}.

answer:

We know their dot product, so we just need to calculate their \htmladdnormallink{magnitudes}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html} $ \left | \mathbf{A} \right | = \sqrt{A_x^2 + A_y^2 + A_z^2} = \sqrt{5^2 + 2^2 +(-1)^2} $ \\ $ \left | \mathbf{A} \right | = \sqrt{25 + 4 + 1} = \sqrt{30} = 5.48 $ \\ \\

$ \left | \mathbf{B} \right | = \sqrt{3^2 + (-3)^2 +(3)^2} $ \\ $ \left | \mathbf{B} \right | = \sqrt{9 + 9 + 9} = \sqrt{27} = 5.2 $ \\ \\

Finally

$ \mathbf{A} \cdot \mathbf{B} = \left | \mathbf{A} \right | \left | \mathbf{B} \right | \cos \theta$

$ \theta = cos^{-1} \left ( \frac{ \mathbf{A} \cdot \mathbf{B} }{ \left | \mathbf{A} \right | \left | \mathbf{B} \right |} \right )$ \\

$ \theta = cos^{-1} \left ( \frac{ 6 }{ (5.48)(5.2)} \right) $ \\ $ \theta = cos^{-1} (0.21) = 77.9^o $

\end{document}