Linear Algebra Exam Help!!

If u=(1,2,1) and v=(-1,1,2)

What are the components of 2u-3v? (I dont know what they mean by components)

Find the orthogonal projection of u and v.

Find the vector component of u orthogonal to v.

Let w = (1,1,0). Determine whether u,v and w lie in the same plane when positioned so that their initial points coincide. (This one is a tuffy)

If you just answer one or two im sure i can figure out the rest based on your answer. Thanks a lot i reallllyyy need help.

I'll respond to the "tuffy"

Let w = (1,1,0)
One quick way to do this problem:
u and v do not lie on the same line, so they generate a plane. If we could show that for some nonzero a,b,c, that au+bv+cw = (0,0,0), we'd be done, because it would tell us that w can be written as a linear combination of u and v, which means in particular that w lies in the plane generated by u and v.
Do this by solving the three equations that you get from the components:
a-b+c = 0
2a+b+c = 0
a+2b = 0
-------------------------
The third equation gives a = -2b
The first two equations then become:
-2b - b + c = 0
-4b + b + c = 0
Both of these equations are the same, and are namely:
-3b + c = 0
c = 3b

So then if we pick a = 2, b = -1, c = -3, we see that
au+bv+cw = (0,0,0)

This tells us that you can write w as a linear combination of u and v; in particular, it tells us that w lies in the plane that u and v generate.

Or you can notice that 2u-v-3w = (0,0,0)
(check this, and make sure you get the same answer using both methods.)

2u - 3v = 2(1,2,1) - 3(-1,1,2) = (2,4,2) + (3,-3,-6) = (2 + 3,4 + (-3),2 + (-6)) = (5,1,-4). The components are 5, 1, and -4. Tough, huh?