No, the magnitude of a vector (in Euclidean space) is the square root of the sum of the squares of its components. This value can never be greater than the value of one of its own components.

v = √((vx)2 + (vy)2 + (vz)2)

v2 = (vx)2 + (vy)2 + (vz)2

(vx)2 = -(vy)2 - (vz)2 + v2

vx = √(-(vy)2 - (vz)2 + v2)

Can vx > v?

Substituting:

√(-(vy)2 - (vz)2 + v2) > √((vx)2 + (vy)2 + (vz)2).

Simplified:

v2 > (vx)2 + 2(vy)2 + 2(vz)2.

Substituting again:

(vx)2 + (vy)2 + (vz)2 > (vx)2 + 2(vy)2 + 2(vz)2.

Simplifying again:

0 > (vy)2 + (vz)2.

This results in a fallacy, since 0 can't be greater than a positive number. This wouldn't work even if both vy and vz were 0.