How do I find 4 linearly independent, perpendicular vectors in a hyper-plane?

Question

Here's the question:

Given the hyper-plane $x+2y-3z-w=0$ in $\mathbb{R}^4$,

a) Find two independent, perpendicular vectors on the plane. What I have done so far is:

$$ [1,2,-3,-1] \cdot [x,y,z,w] = 0 \\ => x=1, y=0, z=0, w=1 $$

So I've only managed to find one linearly independent perpendicular vector.

b) Then find 3 independent, perpendicular vectors.

c) Could you find 4 independent vectors on the hyper plane?

d) Find a fourth vector perpendicular to the previous 3 in the hyperplane. Are perpendicular vectors always independent?

e) What matrix, $A$, has the fourth vector as its null space?

Answer 1

Hint: $$ V = H \oplus H^\top \\ H^\top = \{ t \, n \mid t \in \mathbb{R}, n = (1,2,-3,-1)^T \} $$

Answer 2

Use Gram–Schmidt or Orthogonal-triangular decomposition to find three orthogonal vectors that are on the plane defined by $n=[1\,\,2\,-3\,\,-1]^T$. In MATLAB,

[q,~] = qr([1 2 -3 -1]')

the last three columns of $\mathtt q$ are an answer to the second question. The answer the the third question is no: $n$ is the fourth perpendicular vector. There can only be four perpendicular vectors in $\Bbb R^4$. For the last question, I think $A$ would be the last three columns of $\mathtt q$ with $n$ in the null space.