Four-dimensional vectors?

Question

If you are given three vectors in four-dimensional space and not all are in the same 2-dimensional plane, how would you find a fourth vector perpendicular to these three?

Answer 1

Since we are creatures of only three spatial dimensions, we cannot imagine what a fourth spatial dimension would look like. However, the mathematics are relatively straightforward.

To find a vector that's perpendicular to two others in 3D space, you compute the cross product:

U x V

which is really the determinant of this matrix:

| X^ Y^ Z^ |
| Ux Uy Uz |
| Vx Vy Vz |

where X^ is the x-direction unit vector, and likewise for Y^ and Z^, Ux is the x-component of vector U, etc.

The result is a vector that's perpendicular to both the first two.

To do this in 4D space and find a vector that's perpendicular to vectors U, V, and W, you would simply extend the matrix:

(U x V) x W

which is really the determinant of the matrix

| X^ Y^ Z^ A^ |
| Ux Uy Uz Ua |
| Vx Vy Vz Va |
| Wx Wy Wz Wa |

where X^, Y^, Z^, and A^ are the unit vectors for each dimension, and Wa would be the a-component of vector W.

Answer 2

in case you propose circulate products, then you won't be able to because of the fact the circulate manufactured from 2 vectors is largely a three-d entity. There inspite of the actuality that could desire to be generalisations of this in Clifford Algebras. this is to no longer say you won't be able to multiply 4-D vectors jointly. the main trivial you'll be the dot product, yet i think you be attentive to approximately that already. on the different hand, in case you think of of four-D vectors as quaternions, then that's achieved: i.e, (a,b,c,d) <---> a+bi+cj+dk, the place i^2 = j^2 = ok^2 = ijk = -a million. Multiply those quaternions formally in the time of the distributive components, holding a careful eye on the order of i, j, and ok: as an occasion, ij = -ji. frequently, this multiplication isn't commutative. this is all i will assert on those concerns for now.

Answer 3

system of equations...d:

Edit: Nice insight lithiumd...