I have absolutely no clue. Can someone at least get me started?

Thanks a Million!!!

http://i137.photobucket.com/albums/q...21/Eqn32-1.gif

Printable View

- Jan 21st 2009, 09:02 PMqbkr21Calculus III Word Problem
I have absolutely no clue. Can someone at least get me started?

Thanks a Million!!!

http://i137.photobucket.com/albums/q...21/Eqn32-1.gif - Jan 21st 2009, 10:21 PMGrandadVectors
Hello qbkr21For the geometric argument, consider an arbitrary point in the plane, O, as origin. Then take two lines, $\displaystyle l$ and $\displaystyle m$, through O parallel to $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ respectively (effectively setting up a system of non-perpendicular* coordinate axes). Then consider point C whose displacement from O is represented by $\displaystyle \vec{c}$. From C, draw a line parallel to $\displaystyle l$, meeting $\displaystyle m$ at P. Then $\displaystyle \vec{PC} = s\vec{a}$ for some scalar $\displaystyle s$, and $\displaystyle \vec{OP} = t\vec{b}$ for some scalar $\displaystyle t$, and then:

$\displaystyle \vec{OC} = \vec{OP} + \vec{PC}$

$\displaystyle \Rightarrow \vec{c} = s\vec{a} + t\vec{b}$

For the components argument, let $\displaystyle \vec{a} = a_1\vec{i} + a_2\vec{j}$, and $\displaystyle \vec{b}$ and $\displaystyle \vec{c}$ similarly. Then consider separate components and show that equations

$\displaystyle c_1 = sa_1 + tb_1$

$\displaystyle c_2 = sa_2 + tb_2$

have a solution for some $\displaystyle s$ and $\displaystyle t$.

Grandad

* PS. Not necessarily perpendicular is what I mean. Instead of being covered by a grid of unit squares, as in Cartesian geometry, the plane is spanned by parallelograms with sides $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$.