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Calculus 3 vectors

The question is
If a=<a1,a2> and b=<b1,b2> are nonzero vectors that are not parallel, show that any other vector c in R2 can be written as sa+tb, where s and t are suitably chosen scalers.
So this is a proof and I can solve it geometrically but I would love to solve it algebraically. I found help online where c=a+t(b-a) would give me the answer but I have NO IDEA where c=a+(b-a) came from!
:s-smilie:
Reply 1
Original post by jasonj0118
The question is
If a=<a1,a2> and b=<b1,b2> are nonzero vectors that are not parallel, show that any other vector c in R2 can be written as sa+tb, where s and t are suitably chosen scalers.
So this is a proof and I can solve it geometrically but I would love to solve it algebraically. I found help online where c=a+t(b-a) would give me the answer but I have NO IDEA where c=a+(b-a) came from!
:s-smilie:


1. Let c=(pq)c=\begin{pmatrix} p \\ q \end{pmatrix} and suppose that, as suggested, c=sa+tb\bold{c} = s\bold{a} + t \bold{b}.

2. Express the relationship between the three vectors a,b,c\bold{a},\bold{b},\bold{c} in matrix form.

3. Figure out what you have to be able to do to the matrix that arises in part 2 if you must be able to find the particular ss and tt for any vector c\bold{c}.

4. What relationship must then hold between the components of a\bold{a} and b\bold{b} ?

5. What does that relationship have to do with the non-parallel restriction?
Reply 2
Alternatively:

1) Show that you can express (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1\end{pmatrix} in terms of a\bold{a} and b\bold{b} in the given way.
2) Note that you can express any vector cR\bold{c} \in \mathbb{R} in terms of (01)\begin{pmatrix} 0 \\ 1\end{pmatrix} and (10)\begin{pmatrix} 1 \\ 0\end{pmatrix} in a very obvious way.
3) Substitute your representations for (01)\begin{pmatrix} 0 \\ 1\end{pmatrix} and (10)\begin{pmatrix} 1 \\ 0\end{pmatrix} in terms of a\bold{a} and b\bold{b} to complete the proof.


If you want to go further, you could demonstrate that this representation is unique.

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