# Find the equation of the plane in $\mathbb{ℝ}^{3}$ perpendicular to the subspace $S = \{(\alpha, 3\alpha, -4\alpha):\alpha\in\mathbb{R}\}$

I'm totally lost on how to do this. I know if I was given a normal $$Ax + By + Cz = D$$ plane, I would just have to find the normal vector to a point to find the perpendicular plane, but how do I find the perpendicular plane to a subspace??

• It's perpendicular to $(1,3,-4)$, but there are infinitely many such planes – J. W. Tanner Jun 30 at 2:30
The subspace is $$S={(α,3α,−4α)}$$ which means that it is a trace of the points passing the constant multiples of (1, 3, -4). Meaning the subspace is consisted of the span of one vector, which is (1, 3, -4), so it is just a line passing through the origin with direction vector (1, 3, -4). The rest goes same as you said. $$x+3y-4z=k$$ would be the hyperplane for the vector.(For any real number k)