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Let veca, vecb and vecc be three vectors...

Let `veca, vecb` and `vecc` be three vectors having magnitudes 1,1 and 2 resectively. If `vecaxx(vecaxxvecc)+vecb=vec0` then the acute angel between `veca` and `vecc` is

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The correct Answer is:
`pi//6`
`veca xx (vecaxxvecc) +vecb=vec0`
`or (veca.vecc)veca-(veca.veca)vecc+vecb=vec0`
`or 2 cos theta. veca-vecc+vecb=vec0`
(using `|veca|=1,|vecb|=1,|vecc|=2`)
`or (2costheta veca-vecc)^(2)=(-vecb)^(2)`
`or 4cos^(2)theta.|veca|^(2)+|vecc|^(2)`
`-2.2 cos theta.veca.vecc=|vecb|^(2)`
`or 4cos^(2)theta+4-8costheta.costheta=1`
`or cos^(2)theta-8cos^(2)theta+4=1`
`or 4 cos^(2) theta=3`
`or cos theta = +- sqrt3//2`
for `theta` to be acute , `cos theta= sqrt3/2 Rightarrow theta= pi/6`
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