Find another vector that has the same projection onto v=<1,1> as u=<1,2>.
The vector projection of u onto v
$\displaystyle \frac{u \cdot v}{||v||^2}v=\frac{3}{2}<1,1>$
Now pick an arbitary vector $\displaystyle w=<a,b>$
$\displaystyle \frac{w \cdot v}{||v||^2}v=\frac{a+b}{2}<1,1>$
If you compare these two vectors we see that
$\displaystyle a+b=3 \iff a=3-b$
So the vector is $\displaystyle <3-b,b>$
This will work for any choice of b
P.S go Beavs!!