Vectors in R4 othogonal to other vectors.

[kiwicounter]

1. Find all vecors in R4 orthogonal to both of the vectors (1,2,1,0) and (2,0,3,1)

Homework Equations

a + 2b + c = 0
2a + c + d = 0
d = 4b + c
c = -a -2b

The Attempt at a Solution

a = 1 b = 0

(1,0,-1,3)

a=0 b = 1

(0,1,-2,-2)

Answer: {(1,0,-1,3),(0,1,-2,-2)}

I'm struggling with this stuff, I got this method here, but cannot find the original thread since joining. Thanks for your time.

 

[Mark44]

1. Find all vecors in R4 orthogonal to both of the vectors (1,2,1,0) and (2,0,3,1)

Homework Equations

a + 2b + c = 0
2a + c + d = 0
d = 4b + c
c = -a -2b

The first two equations are the system you need to solve.

I can see where you got the 4th equation, but it's not obvious to me where the 3rd equation came from

Since you have a system of two equations in four unknowns, there are going to be two free variables, meaning that there will be an infinite number of solutions.

The Attempt at a Solution

a = 1 b = 0

(1,0,-1,3)

a=0 b = 1

(0,1,-2,-2)

Answer: {(1,0,-1,3),(0,1,-2,-2)}

Are you saying here that there are just two solutions to the system?

You can check your work, by dotting the vectors you got with the given vectors. <1, 0, -1, 3> works, since it is orthogonal to both of the given vectors, but <0, 1, -2, -2> isn't orthogonal to either of the given vectors.

I'm struggling with this stuff, I got this method here, but cannot find the original thread since joining. Thanks for your time.

 

[kiwicounter]

Thank you. I don't know how to write out the answer for infinite answers.

I got d = 4b + c by substituting c = -a - 2b into the equation a + 2b + c = 0 giving a + 2b +(-a - 2b) = 0, simplifying to 0 = 0 DOH! Mistake.

If I put c = -a -2b into the other equation 2a + c + d = 0...

2a + (-a -2b) +d = 0, simplifies to a - 2b + d = 0 rearranged to d = 2b - a. If this is correct...

I have equations

a + 2b + c = 0
2a + c + d = 0
c = -a -2b
d = 2b-a

a= 1 b = 0

(1,0,-1,-1)

a = 0 b = 1

(0,1,-2,2)

None of my answers are actually orthogonal to both vectors.

I'm lost. I can do a cross product in 3-space but don't know what to do in n-space. Nor do I know how to present an answer that has infinite solutions. I'll persevere, but I am a bit lost. Perhaps sleeping on it will help it's been a long days math for me.

 

[Mark44]

I missed this before, but one of your first two equations has an error. They should be
a + 2b + c = 0
2a + 3[/color]c + d = 0

These equations come from the dot product of your unknown vector <a, b, c, d> with each of your given vectors <1, 2, 1, 0> and <2, 0, 3, 1>.

Do you know how to work with systems in matrix form? The system above can be represented as this 2 x 4 matrix:
[1 2 1 0]
[2 0 3 1]

You can solve the system of equations by row reducing this matrix. If you haven't learned this technique yet, you'll need to work directly with the equations.

Regarding the cross product, it doesn't apply in this problem. The cross product is applicable only to vectors in three-dimensional space.

 

[kiwicounter]

Thanks Mark I really appreciate your help and time here. I'm an ecology major and math is my achilles heel. I'm going to plant a Kauri (giant forest tree) and name it after you to remind me always of the kindness of strangers.

Wow, a mistake right at the start huh. That wouldn't help. I was tired and confused...

If I reduced these in matrix form and solved, would I have an answer resembling 'all vectors in R4 orthogonal to the two vectors given.'

1210
2031 => R2 - 2R1

1 2 1 0
0-4 1 1 => R2 x -1/4 (can I do this?)

1 2 1 0 => R1 - 2R2
0 1 -1/4 -1/4

1 0 1.5 1/2
0 1 -1/4 -1/4

I'm just making a mess I don't know where I'm meant to be going with this sorry.


Try equations. I think what I'm meant to do here is get equations for c and d that do not include c's and d's in those equations - so when arbitrary values are given for a and b, d and c are solvable?

I got c = -a -2b from the first equation then substituted this into the second equation and simplified then rearranged to get the equation d = 2a + 6b for d.

a + 2b + c = 0
a + 3c + d = 0
c = -a - 2b
d = 2a + 6b

a=0 b=1

(0,1,-2,6) This is orthoganal to both.

a=1 b=0

(1,0,-1,2) This didn't work. Only orthogonal to the first vector. I've double checked the original question and it is correct.

So I found one. How do I find 'all of them' without spending an eternity (infinite solutions).

 

[Mark44]

Thanks Mark I really appreciate your help and time here. I'm an ecology major and math is my achilles heel. I'm going to plant a Kauri (giant forest tree) and name it after you to remind me always of the kindness of strangers.

Wow, a mistake right at the start huh. That wouldn't help. I was tired and confused...

If I reduced these in matrix form and solved, would I have an answer resembling 'all vectors in R4 orthogonal to the two vectors given.'

1210
2031 => R2 - 2R1

1 2 1 0
0-4 1 1 => R2 x -1/4 (can I do this?)

1 2 1 0 => R1 - 2R2
0 1 -1/4 -1/4

1 0 1.5 1/2
0 1 -1/4 -1/4

This is exactly right.

Now, let's translate this matrix back into a system of equations involving a, b, c, and d.
a + (3/2)c + (1/2)d = 0
b - (1/4)c - (1/4)d = 0

This also says that
a = -(3/2)c - (1/2)d
b = (1/4)c + (1/4)d
c = c
d = ......d

The last two equations are obviously true.
You can pretty much read off two vectors from the righthand side. You solution is all linear combinations of those two vectors.

In four dimensions, the solution set is a plane.

I'm just making a mess I don't know where I'm meant to be going with this sorry.


Try equations. I think what I'm meant to do here is get equations for c and d that do not include c's and d's in those equations - so when arbitrary values are given for a and b, d and c are solvable?

I got c = -a -2b from the first equation then substituted this into the second equation and simplified then rearranged to get the equation d = 2a + 6b for d.

a + 2b + c = 0
a + 3c + d = 0
c = -a - 2b
d = 2a + 6b

a=0 b=1

(0,1,-2,6) This is orthoganal to both.

a=1 b=0

(1,0,-1,2) This didn't work. Only orthogonal to the first vector. I've double checked the original question and it is correct.

So I found one. How do I find 'all of them' without spending an eternity (infinite solutions).