Finding parallel and perpendicular vectors when |a|=1 and a.b=3?

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Two vectors A and B lie in the same plane and it is given that |A|=1 and A.B=3. I have to find, in terms of A and B, the vector U, parallel to A, and the vector V perpendicular to A. Both of these vectors lie in the same plane such that U+V=A+B.

What I have done: Any vector that is parallel to a will be of the form kA, where k is a real number. Thus I have said U+kA=A+B and so u=(1-k)A + B. For V and A to be perpendicular V.A=0 and so by a rearrangement of the dot product formula v=0(A+B). I am stuck on this problem and I wonder whether anyone could give me some pointers?

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Note that for $k,s,t\in \mathbb{R}$ we have

$$\vec U=k\vec A \quad \vec V=s\vec A+t\vec B$$

since

$$\vec A\cdot \vec V=0\implies s|\vec A|^2+t\vec A\cdot \vec B=0\implies s=-3t \implies \vec V=-3t\vec A+t\vec B$$

then

$$\vec U+\vec V=\vec A+\vec B\implies k\vec A-3t\vec A+t\vec B=\vec A+\vec B\implies t=1 \quad k=4$$

therefore

$$\vec U=4 \vec A \quad \vec V=-3\vec A+\vec B$$