A unit vector perpendicular to both i^−j^ and i^+j^ can be found using the cross product, which gives a vector perpendicular to both input vectors.
The vectors in 3D form:
a=i^−j^+0k^
b=i^+j^+0k^
The cross product:
a×b=i^11j^−11k^00
=i^−1100−j^1100+k^11−11
=i^[(−1)(0)−(0)(1)]−j^[(1)(0)−(0)(1)]+k^[(1)(1)−(−1)(1)]
=i^(0)−j^(0)+k^(2)
=2k^
The magnitude of 2k^:
∣a×b∣=02+02+22
=2
The unit vector:
n^=22k^
=k^
Both given vectors lie in the xy-plane. Any vector perpendicular to this plane points in the z-direction.
Therefore, the unit vector perpendicular to both given vectors is k^.