Hi , not sure if vectors falls under this category , but what are you going to do I just need someone to explain this question for me step by step if you have time . I'm Gavin quite a hard time with it :/

"The vectors a and b span two space . For what values of m is it true that (m^2+2m-3)a + (m^2+m-6) = 0
*note that it is a 0 vector. I imagine this'd and they have to cancel each other out some how but I don't know how ....

Thanks !
Emma

Hey elittlewood.

Did you mean (m^2+2m-3)*a + (m^2+m-6)*b = 0?

If so try setting up equations involving m and each of the vector components (i.e. x, y, and z) and then solve a set of equations.

Okay... woould that mean you have to expand...? im having a hard tie with that .

Basically you have x,y,z for a and another for b.

You will get three sets of three equations. For example if we look only at x co-ordinate, we get:

(m^2 + 2m - 3)*x_a - (m^2 + m - 6)x_b = 0.

You have two other equations involving the other co-ordinates so if you set x_a and x_b to constants you can solve for m and if m is a common solution for all co-ordinates, then you have a solution.

Originally Posted by elittlewood
"The vectors a and b span two space . For what values of m is it true that (m^2+2m-3)a + (m^2+m-6) = 0
Originally Posted by chiro
Basically you have x,y,z for a and another for b.
Please note that there is no second b in the original post. I think there is a typo in the question itself.

I have seen this sort of question before. I think it should be $\displaystyle (m^2+2m-3)\vec{a}+(m^2+m-6)\vec{b}=0$.

Because $\displaystyle \vec{a}~\&~\vec{b}$ span 2-space then $\displaystyle \alpha\vec{a}+\beta\vec{b}=\vec{0}\text{ if snd only if }\alpha=\beta=0$.

Thus coordinates really have little to do with it.

set both $\displaystyle m^2 + 2m - 3$ $\displaystyle \space$ and$\displaystyle \space$ $\displaystyle m^2 + m - 6$ $\displaystyle \space$ equal to $\displaystyle \space$$\displaystyle 0$
solve both quadratics and then what ever value of $\displaystyle m$ is common to both would be the value that you are looking for