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

Explanation

`|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`