# One final question in algebra and geometry!

1. Oct 23, 2005

### byronsakic

the vectors u and v are linearly independent. find s, if vectors (1-s)u - 2/3v and 3u +sv are parallel.
i got them equal to each other....but then if you draw them, both of them would be in the exact place which makes them collinear....so would you say it is parallel because vectors u and v are linearly independent?
this is my solution.

2. Oct 23, 2005

### Gokul43201

Staff Emeritus
You can't have two different values of s (one for u and one for v).

Also, two parallel vectors need not be equal.

If A is parallel to B (A,B vectors), then A = kB (k is a scalar).

Last edited: Oct 24, 2005
3. Oct 23, 2005

### byronsakic

any further suggestions?

like this is a question we've never seen before so i don't know how to approach the question, using a scalar a = kb with the s would give me k and s right. but i have to solve for s.

Last edited: Oct 23, 2005
4. Oct 24, 2005

### HallsofIvy

You know that {(1-s)u- (2/3)v}= k{3u+ sv} AND that u and v are independent.

{(1-s)- 3k}u- {(2/3)+sk}v= 0. Since u and v are indepent, you must have

1-s- 3k= 0 and (2/3)+ sk= 0, two equation in two unknowns. Since one of the equations is non-linear, there may be two correct solutions. However, the same value of s applies to both the u and v equations.

5. Oct 24, 2005

### byronsakic

thanks a lot :D