Now, linking end to end 10 vectors $\overrightarrow{\mathrm{AB}}$ comes to multiplying the vector $\overrightarrow{\mathrm{AB}}$ by 10, so that one obtains a vector with the origin at (0,0,0) and end point at (10,10,10).

For the vector $\overrightarrow{\mathrm{CD}}$: $\overrightarrow{\mathrm{CD}}=i+3j+2k$. Adding up 2 such vectors, one easily sees that the resulting vector has the origin at (0,0,0) and the end point at (2,6,4). Adding up 10 such vectors, the origin remains the same and the end point is (10, 30, 20). For $n$ vectors, the end point is $(n,3n,2n)$.

If the starting point is (1,0,-1) and the vector to be drawn is $v=ni+nj+nk$, then the end point will be at $(1+n,0+n,-1+n)$, i.e. at $(n+1,n,n-1)$.

If the origin is at $(a,b,c)$, the end point is at $(n+a,n+b,n+c)$. If the vector is $u=ni+3nj+2nk$, and its origin is at (-3,0,2), its end point will be at $(n-3,3n,2n+2)$. If for the same vector the origin is at $(a,b,c)$, its end point will be at $(n+a,3n+b,2n+c)$.

If the vector of interest is now $u+v=2ni+4nj+3nk$, in the general case of the origin at $(a,b,c)$, the end point will be at $(2n+a,4n+b,3n+c)$.

Stephen (from Framwellgate School, Durham) solved the problem and went on to think about what happens if you alternate segments. Here is what he sent us :

With AB followed by CD (if we count AB as segment 1 and CD as segment 2) we can see that 2 segments makes (1,1,1)+(1,3,2)=(2,4,3) therefore the nth segment when n is even is (n,2n,3n/2) because we need to divide it by 2, as (2,4,3) is 2 segments. For n segments when n is odd, n-1 is even so take the n-1th segment from last coordinates to give (n-1,2(n-1),3(n-1)/2) and add (1,1,1), because this is the first segment and therefore will be the last segment on odd number of segments (n) - to get: (n,2n-1,(3n-1)/2) if starting from (a,b,c) then we just need to add this coordinate to the two solutions.