Scalar Product is defined as ##\mathbf A \cdot \mathbf B = | \vec A | | \vec B | \cos \theta##.(adsbygoogle = window.adsbygoogle || []).push({});

With the construct of a triangle, the Law of Cosines is proved.

##\mathbf A## points to the tail of ##\mathbf B##.

Well, ##\mathbf C## starts from the tail of ##\mathbf A## and points to somewhere.

Finally, ##\mathbf B## points to the head of ##\mathbf C##.

Now someone defines a new operation such as ##\mathbf A \circ \mathbf B = | \vec A | | \vec B | \sin \theta##. Following the same steps of proving the Law of Cosines.

##\begin{align}

\mathbf C & = \mathbf A + \mathbf B \nonumber \\

\mathbf C \circ \mathbf C & = (\mathbf A + \mathbf B) \circ (\mathbf A + \mathbf B) \nonumber \\

| \vec C | | \vec C | \sin 0 & = | \vec A | | \vec A | \sin 0 + | \vec B | | \vec B | \sin 0 + 2 | \vec A | | \vec B | \sin \theta \nonumber \\

| \vec A | | \vec B | \sin \theta & = 0 \nonumber

\end{align}##

##\theta = 0##, ##\mathbf A = 0## or ##\mathbf B = 0##?

I am not quite understand what the result means. It seems the result is not consistent with the constructed triangle.

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# I Scalar Product

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