Is a Vector Orthogonal to a Timelike Vector Necessarily Spacelike?

[kontejnjer]

Homework Statement

Show that if x^{\mu} is timelike and x^{\mu}y_{\mu}=0, y^{\mu}\neq 0, then y^\mu is spacelike.

Homework Equations

ds^2=\\>0\hspace{0.5cm}\text{timelike}\\<0\hspace{0.5cm}\text{spacelike}\\0\hspace{0.5cm}\text{lightlike}
metric is diag (+---)

The Attempt at a Solution

Don't know if this is the correct way, but here goes: assuming that x^\mu is timelike, we can pick a reference frame in which x^\mu=(x^0,\vec{0}), so due to invariance x^{\mu}y_{\mu}=x^0 y_0=0\rightarrow y^0=0, but, since it is stated that y^{\mu}\neq 0, then \vec{y}\neq 0, and hence we have y^\mu y_\mu=y^i y_i=-(\vec y)^2<0, making y^\mu indeed spacelike. Am I missing something here or is this the right procedure?

 

[Orodruin]

Your procedure is fine.