# Linear Combination Method

1. Jan 4, 2008

### Mtl

[SOLVED] Linear Combination Method

1. The problem statement, all variables and given/known data
Show, using the linear combination method, that the vectors below are non-coplanar or independent. Be complete
d= [2,-1,-2], e=[1,1,1] , and f = [1,-5,8]

2. Relevant equations
Ok so I'm assuming, I am supposed to use Cramer's Rule here. But the problem is I don't fully understand it.

3. The attempt at a solution

2. Jan 4, 2008

### Shooting Star

Just find the magnitude of the determinant.

3. Jan 5, 2008

### Defennder

Why do you have to use Cramer's rule? There are many ways to do the question. One way, as Shooting star has pointed out, but not fully enough is to treat the three vectors as row vectors in a 3x3 matrix. Now what do you remember of the properties of a matrix whose row vectors are linearly independent?

4. Jan 5, 2008

### HallsofIvy

Staff Emeritus
What, exactly do you mean by "the linear combination method"? I would think that it just means to use the definition of "independent"- that the only way a linear combination of the vectors can be 0 is if the coefficients are all 0: Show that
$\alpha [2, 1, -2]+ \beta [1, 1, 1]+ \gamma [1, -5, 8]= [0, 0, 0]$ only if $\alpha= \beta= \gamma= 0$.

5. Jan 5, 2008

### Shooting Star

Why not take the shortcut and arrange the three vectors as three rows and show that the det not equal to 0? He will be finding the det anyway to solve the eqns, if he uses Kramer's rule.

6. Jan 5, 2008

### Mtl

OK, thanks for your input, I think I have it now.

7. Jan 6, 2008

### Mtl

New Question... Given that the two vectors u=(2a+b) and v=(4a-3b) are perpendicular and that |a|=3 and |b|=6, then find the angle between a and b.

So I pretty sure i need to set the dot product of u and v = to zero. Then there is probably some sort of way to rearange it, of which I am not sure.

8. Jan 6, 2008

### Shooting Star

Just take the scalar product of u and v, which is zero. You'll get terms containing a^2, b^2 and a.b. So, find cos theta.

9. Jan 6, 2008