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Prove that : (vecaxxvecb)^2=|\veca|^2.|\...

Prove that : `(vecaxxvecb)^2=|\veca|^2.|\vecb|^2-(veca.vecb)^2`.

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If veca and vecb are any two vectors, prove that (vecaxxvecb)^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2) It is known as Largange's indentity.

Prove that |vecaxxvecb|^2=|veca|^2|vecb|^2-(veca*vecb)^2=|:[veca*veca veca*vecb],[veca*vecb vecb*vecb]:| .

If veca" and "vecb are any two vectors, then (veca xx vecb)^(2)= |veca|^(2)|vecb|^(2)-(veca* vecb)^(2) .

If veca and vecb are two vectors , then prove that (vecaxxvecb)^(2)=|{:(veca.veca" ",veca.vecb),(vecb.veca" ",vecb.vecb):}|

Show that (veca-vecb)xx(veca+vecb) = 2(vecaxxvecb)

Show that (veca-vecb)xx(veca+vecb)=2(vecaxxvecb) .

Prove that (veca+vecb) cdot (veca+vecb) = |veca|^2 + |vecb|^2 , if and only if veca, vecb are perpendicular, given veca ne vec0, vecb ne vec0 .

If vecb and vecc are two non-collinear such that veca ||(vecbxxvecc) . Then prove that (vecaxxvecb).(vecaxxvecc) is equal to |veca|^(2)(vecb.vecc)

JMD PUBLICATION- A Gift paper from the publishers-Example
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