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.