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In isosceles triangle ABC |vec(AB)|=|vec...

In isosceles triangle ABC `|vec(AB)|=|vec(BC)|=8` a point E divides AB internally in the ratio 1:3, then find the angle between `vec(CE) and vec(CA)` ( where `|vec(CA)|= 12`)

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`|vecc|=12and |vecb|=|vecb-vecc|=8`
`b^2=b^(2)+c^(2)-vecb.bvecc`
`vecb.vecc=72`
`costheta(vecc.(vecc-((vecb)/(4))))/(|vecc||vecc-vecb/4|)=(vecc.vecc-(vecc.vecb)/4)/(12|vecc-vecb/4|)=(144-18)/(12|vecc-vecb/4|)`
`|vecc-vecb/4|^(-2)=|vecc|^(2)+|vecb|^(2)/16-(vecb.vecc)/2=144+4-36=112`
`cos theta=21/(2xxsqrt112)=21/(2xx4sqrt7)=(3sqrt7)/8`
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