# Linear algebra

1. Dec 5, 2013

### sevag00

1. The problem statement, all variables and given/known data

Find two vectors in R4 of norm 1 that are orthogonal to the vectors u = (2, 1, −4, 0),
v = (−1, −1, 2, 2) and w = (3, 2, 5, 4).

2. Relevant equations

3. The attempt at a solution

What i did was, i let a vector x = (x1, x2, x3, x4) that has a norm of 1 and orthogonal to u,v and w
i.e. x.u = 0; x.v = 0 and x.w = 0.
This where i'm stuck. Some help would be appreciated. Thanks.

2. Dec 5, 2013

### tiny-tim

hi sevag00!

(try using the X2 button just above the Reply box )

hint: how would you find a vector (x1,x2,x3) orthogonal to (a,b,c) and (d,e,f) ?

3. Dec 5, 2013

### sevag00

Using dot product. x1.a + x2.b + x3.c = 0. Same for (d, e, f).

4. Dec 5, 2013

5. Dec 5, 2013

### sevag00

Yeah, but the chapter only explains the dot product. So, i think the problem should be solved using dot product.

Last edited: Dec 5, 2013
6. Dec 5, 2013

### tiny-tim

ok, in that case you have three simultaneous equations (from the dot products) …

solve them the usual way

7. Dec 5, 2013

### sevag00

3 equations with 4 unknowns. Will this yield a solution?

8. Dec 5, 2013

### tiny-tim

3 equations with 3 unknown ratios

so yes, you can find the ratios

9. Dec 5, 2013

### sevag00

Ah yes. 4 equations with 4 unknowns.

10. Dec 5, 2013

### HallsofIvy

Further, the cross product is defined only in three dimensions while this problem is in four dimensions.

11. Dec 5, 2013

### sevag00

Wait a minute. You got me confused. There are 3 equations with 4 unknown. The equation of the norm cannot be included with the other three.

12. Dec 5, 2013

### tiny-tim

i was talking about three dimensions
there are only three unknown ratios

so you can find the ratios

13. Dec 5, 2013

### sevag00

What ratios? I want to find the components of x.

14. Dec 5, 2013

### tiny-tim

x2/x1, x3/x1, x4/x1

15. Dec 5, 2013

### sevag00

16. Dec 5, 2013

### tiny-tim

divide your three dot-product=0 equations by x1

that gives you three equations in x2/x1, x3/x1, and x4/x1

then solve them

17. Dec 5, 2013

### sevag00

I think i did it right. I replaced x2/x1 with X x3/x1 with Y and x4/x1 with Z. Then solved 3 equations with 3 unknowns.

18. Dec 5, 2013

### Ray Vickson

You can write the three equations in four unknowns as three equations in three unknowns but with the fourth unknown on the right-hand-side. In other words, you can express three of the x_i as functions of the fourth. For example, you can write the second equation $-1 x_1 - 1 x_2 + 2 x_3 + 2 x_4 = 0$ as $-1 x_1 - 1 x_2 + 2 x_3 = -2 x_4$, or $x_1 + x_2 - 2x_3 = 2x_4$. Do the same thing for the third equation; the first already has the right form. Now you can solve for $x_1, x_2, x_3$; this will give them as formulas involving $x_4$.

BTW: the method above is a standard way of dealing with such problems; it is used all the time in such fields as linear optimization (linear programming).

19. Dec 5, 2013

### sevag00

I know that method. But then those equations will become parameters. They will depend on the value of x4 which has infinite solutions.

20. Dec 5, 2013

### Ray Vickson

Of course; that is the whole point. So, if you want to find two different solutions, just give two different values for $x_4$---any two values you like, except 0. (Putting $x_4 = 0$ gives the zero vector, which cannot be normalized to length = 1!)