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Dot product arithmetic |
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| Jan14-13, 06:39 PM | #1 |
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Dot product arithmetic
If you square the magnitude of a vector you get the dot product, correct?
||v||^2 = v . v Can you also say that ||v|| = sqrt(v . v)? |
| Jan14-13, 07:05 PM | #2 |
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Of course, basic algebra.
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| Jan14-13, 07:09 PM | #3 |
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| Jan14-13, 08:07 PM | #4 |
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Dot product arithmetic
Okay. Just to cement this:
If ##\vec{v} = <a,b>## and ##\vec{w} = <c,d>## then ##\vec{v} \cdot \vec{w} = ac+bd## and ##\vec{v} \cdot \vec{v} = a^2+b^2## So if ##|| \vec{v} || ^2 = \vec{v} \cdot \vec{v} = a^2+b^2## then ##\sqrt{|| \vec{v} || ^2} = || \vec{v} || = \sqrt{a^2+b^2}## |
| Jan14-13, 10:11 PM | #5 |
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Hey cytochrome.
If the inner product is valid then all of your statements are true. |
| Jan15-13, 04:23 AM | #6 |
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