Linear Algebra - Equation for a Line Through 2 Points

[Throwback]

Homework Statement

In Rⁿ, define A = (0, 0, ..., 0) and B = (1, -1, ..., (-1)^n-1)

Find (1) the parametric form of the line through A and B, (2) as an intersection of (n-1) hyperplanes, and (3) the hyperplane crossing the origin, normal to AB.

Homework Equations

AB = ((1-0),(-1-0),...,(-1)n-1-0) = (1,-1,...,(-1)^n-1)

The Attempt at a Solution

A = (0,0,...,0)
B = (1,-1,...,(-1)^n-1)

AB = ((1-0),(-1-0),...,(-1)^n-1-0)
AB = (1,-1,(-1)^n-1)

Parametric form:

[x₁ x₂ ... x_n] = [0 0 ... 0] + s[1 -1 ... (-1)^n-1]

Intersection with plane (I'm really lost here):

a₁(x₁)+a₂(x₂)+...+a_n(x_n)
a₁(0+1s)+a₂(0-1s)+...+a_n(0+(-1)^n-1s)
a₁(s)-a₂(s)+...+a_n((-1)^n-1s)

Hyperplane crossing the origin, normal to AB:

Would this just be the n_AB x n_plane?

 

[Dick]

In general the equation of a hyperplane through the origin is N.x=0 (where N is the normal to the plane, x is a general point on the plane and '.' means dot product), right? In part (3) they are asking for a plane that's normal to the vector AB. What would that be? For part (2) you want n-1 different planes that contain your line. I'll give you a hint. Put N=(1,1,0,...,0). Does that contain your line? Can you give me n-2 more?

 

[Throwback]

I'm still having some trouble picturing this...

So for part (3), I need to find a plane that's normal to vector AB.

Let n be my normal vector, AB. I now have to calculate n.p, where p is some point that lies in the plane. I'll use A for my point since it wants it at the origin and A is all zeros. I then have the following:

1*0+(-1)*0+... which is equal to 0.

For a plane, I'd typically use ax+by+cz=d in R³. Because this is Rⁿ, I assume this extends out through n? Since I solved for d via n.B, d=0. The coefficients in the plane equation will be equal to my normal vector values. My plane is then (hopefully):

1x₁-1x₂+...+(-1)^n-1x_n=0
x₁-x₂+...+(-1)^n-1x_n=0

Is that my plane in general form?

Now, again, I'm intimidated by (2). I'm not quite sure where you got that N, Dick. Is is because N (the vector AB) is normal to the plane. We want other hyperplanes that intersect said hyperplane, and then cannot be normal, otherwise they'd be parallel. So we choose a different N?

 

[Dick]

I'm still having some trouble picturing this...

So for part (3), I need to find a plane that's normal to vector AB.

Let n be my normal vector, AB. I now have to calculate n.p, where p is some point that lies in the plane. I'll use A for my point since it wants it at the origin and A is all zeros. I then have the following:

1*0+(-1)*0+... which is equal to 0.

For a plane, I'd typically use ax+by+cz=d in R³. Because this is Rⁿ, I assume this extends out through n? Since I solved for d via n.B, d=0. The coefficients in the plane equation will be equal to my normal vector values. My plane is then (hopefully):

1x₁-1x₂+...+(-1)^n-1x_n=0
x₁-x₂+...+(-1)^n-1x_n=0

Is that my plane in general form?

Now, again, I'm intimidated by (2). I'm not quite sure where you got that n, Dick.

For part (3) it may just be that the answer is so simple it's confusing you. AB.x=0 is a plane normal to the direction of vector AB, isn't it? There's really nothing to check. For part (2) your line is [s,-s,s,-s,...,(-1)^(n-1)s], right? I picked the vector N=[1,1,0,0,..0] because I noticed that N.[s,-s,s,-s,...,(-1)^(n-1)s]=s-s=0. So all points on the line are normal to N. So all points on the line are in the plane normal to N. Now give me some more normals.

 

[Throwback]

N₃ = (0,0,1,1,0,0,...,0)

N₄ = (0,0,0,0,1,1,...,0)

Yeah?

Because these would result in the following dot products (with AB):

(0+0+1-1+0+0+...+0)=0, thus normal
(0+0+0+0+1-1+0+0+...+0)=0, thus normal

 

[Dick]

N₃ = (0,0,1,1,0,0,...,0)

N₄ = (0,0,0,0,1,1,...,0)

Yeah?

Because these would result in the following dot products (with AB):

(0+0+1-1+0+0+...+0)=0, thus normal
(0+0+0+0+1-1+0+0+...+0)=0, thus normal

True, but that's only going to give you about n/2 normals, right? You want n-1. What about (0,1,1,0,0,...0)?

 

[Throwback]

Yeah, because I was shifting over the pair of ones by two every time, and hence decreased by n/2 total normals -- makes sense. I'm not quite sure why I did that; probably just didn't think about the easy math clearly enough.

So, just shifting the pair of twos will result in a different normal vector, and for n-1, we'd have (0,1,1,0,...0), and n-2 (0,0,1,1,0,...,0), etc.

 

[Dick]

Yeah, because I was shifting over the pair of ones by two every time, and hence decreased by n/2 total normals -- makes sense. I'm not quite sure why I did that; probably just didn't think about the easy math clearly enough.

So, just shifting the pair of twos will result in a different normal vector, and for n-1, we'd have (0,1,1,0,...0), and n-2 (0,0,1,1,0,...,0), etc.

Ok! So that gives you n-1 hyperplanes that contain your line. Now can you prove (if you have to, you might just have to show them without proving it, I'll leave that call up to you) that the ONLY points in the intersection of all of those n-1 hyperplanes must lie on your line?