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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??

Thanks in advance for any advice/help!

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  • $\begingroup$ It's perpendicular to $(1,3,-4)$, but there are infinitely many such planes $\endgroup$ – J. W. Tanner Jun 30 at 2:30
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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)

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  • $\begingroup$ Oh I see, so it's pretty much the same process as usual. The \alpha in the subspace got me scared...could you elaborate more on why it isn't included in your answer? $\endgroup$ – Qing Zheng Jun 30 at 2:39

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