# 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.

Thankyou.

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

If I have

$x^2+y^2+z^2+w^2 = R$

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.
Are there any good resources for this kind of topic? If I wanted to buy a book on this stuff it seems more specialised than calculus, what is this kind of topic called?
Thanks.