Question

How to find cross product of 2 vectors in R^4 ?

Answer 1

\(\left[\begin{matrix} 1\\1\\0\\0\end{matrix}\right]\) and \(\left[\begin{matrix} -1\\0\\1\\0\end{matrix}\right]\)

Answer 2

Let v is the required vector, so that v orthogonal to both vector above. It is easy to see that \(v= \left[\begin{matrix} 0\\0\\0\\1\end{matrix}\right]\) satisfies the condition. But I don't want it. I want the method to find it out, not just trial and error. Please, help me out

Answer 3

as far as i can recollect, cross product is only defined in R^3

but I do recall a paper ina journal that discussed the usefulness of a nxm determinant where n /= m

Answer 4

So, how can we find it out in this case?

Answer 5

we can use dot product also. right?
Let v =<a, b, c, d> , hence v . v1 = < a, b, c, d> .< 1,1,0,0> = 0 (we need it)
and v. v2 = < a, b, c, d> .< -1,0,1,0> =0 also.
But it is hard to find out the result

Answer 6

yes, dotproduct will determine if a vector is parralel to another.

Answer 7

for the first one, it is a + b =0
the second one , it is -a + c = 0
hence a = -b, and c, d are arbitrary,
a = c and b, d are arbitrary
then v can be < 1, -1, 1, 0> , but it doesn't satisfy v.v1 =0 or v.v2 =0 :(

Answer 8

the plane in R^4 is most likely teh sae approach as a circle to a sphere, add on the missing dimension

Answer 9

I know my logic is flaw at somewhere, but don't know how to fix yet

Answer 10

n = (a,b,c,d)

a(x-xo) + b(y-yo) + c(z-zo) + d(ko) = 0

but then im thinking that in R^4 there are multiple 'normals' to a plane

Answer 11

k-ko that is

Answer 12

oh, i got it.
the first one gives me <a, -a, 1, 1> since I can pick any of c, d

the second one gives me <a, 1, a , 1> since i can pick any of b and d
But they must be the same. that is < a, -a, 1, 1> =< a, 1, a, 1> hence a = a, -a =1 , 1 = a.
it is possible if and only if a =0, hence the required v is v =< 0,0,0,1> Am I right?

Answer 13

abcd
1100
-----
a+b = 0

abcd
-1010
------
-a+c= 0


a = a
b = -a
c = a
d = d