Orthogonal and Parallel Vectors

[_N3WTON_]

Homework Statement

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}

Homework Equations

Dot product
Cross product

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.

 

[Dick]

Homework Statement

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}

Homework Equations

Dot product
Cross product

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.

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.

 

[_N3WTON_]

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.

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

 

[Dick]

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

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?

 

[_N3WTON_]

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?

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?

 

[Dick]

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?

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

 

[_N3WTON_]

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

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

 

[_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 ?

 

[Dick]

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 ?

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

 

[_N3WTON_]

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

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

 

[_N3WTON_]

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

 

[_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

 

[Dick]

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

Yes, c=1/2 not 2.

 

[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.

 

[_N3WTON_]

Yes, c=1/2 not 2.

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

 

[_N3WTON_]

sd

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.

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

 

[WWGD]

Sure, there are many versions here; let me know if you need a followup.

http://www.cuil.pt/r.php?cx=0028257...ndamental+Theorem+of+Linear+Algebra&sa=Search

 

[_N3WTON_]

Sure, there are many versions here; let me know if you need a followup.

http://www.cuil.pt/r.php?cx=002825717068136152164:qf0jmwd8jku&cof=FORID:10&ie=UTF-8&q=Fundamental Theorem of Linear Algebra&sa=Search

thanks, I'm about to go to bed but ill look over that tomorrow and let you know if I do need a followup