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Prove that |axxb|^2 =a^2b^2 - (a.b)^2...

Prove that `|axxb|^2 =a^2b^2 - (a.b)^2`

Text Solution

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Let `|a| = a,|b|=b` and `theta` be the angle between them.
`|axxb|^2 = (ab sin theta)^2 = a^2b^2 sin^2 theta`
`=a^2b^2(1-cos^2 theta) = a^2b^2 - (a.b cos theta)^(2)`
`=a^2b^2 - (a.b)^2`
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