# Thread: Finding a vector's coordinates in 4 space which is normal to a plane

1. ## Finding a vector's coordinates in 4 space which is normal to a plane

If I have a plane

aw + bx + cy + dz + k = 0

Isn't the normal vector just <a,b,c,d> ? I know this concept is true in three-space, so I'm thinking it naturally extends to four-space, and on and on really? Is that reasonable?
If i was asked to find the coordinates of the normal vector? is that the same as finding the normal vector?

2. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

The "normal" vector in higher dimensions is frequently called an orthogonal vector. You have a vector in the plane (say $\left< w,x,y,z\right>$). That vector should be orthogonal to some vector $\left<w',x',y',z'\right>$. Your guess is, one such vector is $\left<w',x',y',z'\right> = \left<a,b,c,d\right>$. Two vectors are orthogonal if their dot product is zero. So, let's test your theory:

$aw+bx+cy+dz = -k$

They are orthogonal if and only if $k=0$.

3. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

No, SlipEternal is not quite correct. If k is not 0, then the plane does not contain the origin. If $(w_0, x_0, y_0, z_0)$ is a specific point lying in the plane and $(w, x, y, z)$ is generic point in that plane then $$ is a vector lying in that plane so that $a(w- w_0)+ b(x- x_0)+ c(y- y_0)+ d(z- z_0)= aw+ bx+ cy+ dz- (aw_0)+ b(x_0)+ c(y- y_0)+ d(z- z_0)a= k- k= 0$.

Yes, <a, b, c, d> is normal to the plane aw+ bx+ cy+ dz= k and the same thing extends to any dimension.

5. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

OH that's good thanks Halls. Yes it might have been my notation that confused you there Slip.
Thanks both of you.

6. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

I have another question related to this, while we are on this topic.
Based on the hyperplane

$ax+by+cz+dw = k$

Where $a, b, c, d$ and $k$ are real and fixed numbers.

We have two hyperplanes $x = 1$ and $y = 2$. I have to find the equation of the intersection of that and also describe what it is geometrically. Well I'm pretty sure it's a plane but I think there is more to be added to the worded description of what it is, I also have to get the equation of this intersection.

I was thinking to solve the two equations

$a+by+cz+dw = k$
$ax+2b+cz+dw = k$

but at the same time, I think my lack of understanding is leading to a more eloquent and succinct solution.
I mean if we talk about a 3d space, and are talking about two planes $x=1$ and $y=2$ and we have to describe and find the equation of the intersection of these two planes, well, I could easily describe this as a line the line (1,2, z) I believe.
The equation of it would be the parametric equations: $x = 1, y = 2, z = t$. or maybe even the vector $\vec{r} = <1,2,t>$

Anyone got any input to this?

7. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

Also is the normal vector <a,b,c,d> relevant to the question in the previous post?
$ax+by+cz+dw = k$

8. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

No, those planes are NOT described by equations a+ by+ cz+ dw= k and ax+ 2b+ cz+ dw= k. They are described by the equations given: x= 1 (a= 1, b= c= d= 0, k= 1) and y= 2 (b= 1, a= c= d= 0, k= 2). Their intersection is the plane given by the equation a+ 2b+ cz+ dw= k or cz+ dw= k- a- 2b.

9. ## Re: Finding a vector's coordinates in 4 space which is normal to a plane

$x^2+y^2+z^2+w^2 = R$
and I have a hyperplane $x=r$ and $r=R$
I'm expecting that to be a single point right, if not, why not? Also, what about when $r>R$? I'm thinking in this case that there is no intersection. It's very hard to visualise or even conceptualise in these higher dimensions.