# Orthogonal and Parallel Vectors

1. Oct 4, 2014

### _N3WTON_

1. The problem statement, all variables and given/known data
For the following vectors 'u' and 'v', express 'u' as the sum u = p + n where 'p' is parallel to 'v' and 'n' is orthogonal to 'v'
u = {-1, 2, 3}
v = {2, 1, 1}

2. Relevant equations
Dot product
Cross product

3. The attempt at a solution
First, I should say that I do not know how to use latex for determinants, so if somebody can help me out there I could make this problem a bit easier to read. Now, I am aware that two vectors are orthogonal if their dot product is 0 and parallel if their cross product is zero. I used this knowledge to attempt to solve the problem:
0 = p x v
0 = n 'dot' v
To attempt to figure out the value of p={a, b, c} I performed the cross product and obtained:
$i (c-b) - j (2c-a) + k (2b-a) = 0$
$c = b$
$2c = a$
$2b = a$
Next, to attempt to figure out the value of n={x, y, z} I performed a the dot product and obtained:
$(2, 1, 1) dot (x, y, z)$
$2x + y + z = 0$
At this point I am not sure what to do, if somebody could give me some advice I'd greatly appreciate it.

2. Oct 4, 2014

### Dick

You are thinking too hard. cv is parallel to v for any choice of scalar c. Let that be parallel part. Then the other part of the sum must be u-cv. Find c by making sure u-cv is perpendicular to v using the dot product.

3. Oct 4, 2014

### _N3WTON_

Ok so I made c =2:
cv = {4, 2, 2} = p
n = u - p = {-1, 2, 3} - {4, 2, 2} = {-5, 0 , 2}
u = p + n = {4, 2, 2} + {-5, 0 ,2} = {-1, 2, 4}
is the problem really this simple? XD

4. Oct 4, 2014

### Dick

Yes, it's basically that simple. But I don't get c=2. Your n isn't perpendicular to v. How did you get that?

5. Oct 4, 2014

### _N3WTON_

It was arbitrary, I thought two vectors are parallel so long as their components are proportional, {4, 2, 2} is proportional to {2, 1, 1}, no?

6. Oct 4, 2014

### Dick

Yes, but u-cv needs to be perpendicular to v as well. That's what determines c.

7. Oct 4, 2014

### _N3WTON_

ohhh...ok, let me try again..

8. Oct 4, 2014

### _N3WTON_

nvm I thought I knew how to do it but I dont....would I find c by:
{-1, 2, 3} - c{2, 1, 1} = 0 ?

9. Oct 4, 2014

### Dick

No. u-cv is supposed to perpendicular to v. Use the dot product! So (u-cv).v=0. Solve that.

10. Oct 4, 2014

### _N3WTON_

sorry I thought I solved it but made a mistake....trying again

11. Oct 4, 2014

### _N3WTON_

Ok I got c=1/2, is that correct?

12. Oct 4, 2014

### _N3WTON_

sorry, I should show how I got that....
(u-cv).v = 0
u.v - cv.v = 0
{-1,2,3}.{2,1,1} - c{2,1,1}.{2,1,1} = 0
3 - 6c = 0
c = 1/2

13. Oct 4, 2014

### Dick

Yes, c=1/2 not 2.

14. Oct 4, 2014

### WWGD

Just to state, about the general case, that in any vector space V where orthogonality makes sense and W is a subspace of V , we have $V= W \oplus W^{\perp}$ then every vector can be written as the sum of a vector w in W and a vector $w^{\perp}$ in $W^{\perp}$. This also follows from the Fundamental Theorem of Linear Algebra.

15. Oct 4, 2014

### _N3WTON_

thank you for your help...youre right, i was thinking to much about that one XD

16. Oct 4, 2014

### _N3WTON_

sd
I appreciate your help but I'm not familiar with the FToLA (perhaps you could elaborate)....I plan on taking Linear Algebra next semester :D

17. Oct 4, 2014

### WWGD

18. Oct 4, 2014