For this question, to find the orthogonal complement for $\operatorname{sp}([1,3,0],[2,1,4])$,do I just take the nullspace $Ax=0$? I am not asking for the answer, I just want to know if I have the right approach.
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$\begingroup$Orthogonal complement is nothing but finding a basis.
$$\mbox{Let us consider} A=Sp\begin{bmatrix} 1 \\ 3 \\ 0 \end{bmatrix},\begin{bmatrix} 2 \\ 1 \\ 4 \end{bmatrix}$$ $$A^T=\begin{bmatrix} 1 & 3 & 0 & 0\\ 2 & 1 & 4 & 0\end{bmatrix}_{R_1<->R_2}$$ $$=\begin{bmatrix} 2 & 1 & 4 & 0\\ 1 & 3 & 0 & 0\end{bmatrix}_{R_1->R_1\times\frac{1}{2}}$$ $$=\begin{bmatrix} 1 & \dfrac { 1 }{ 2 } & 2 & 0 \\ 1 & 3 & 0 & 0 \end{bmatrix}_{R_2->R_2-R_1}$$ $$=\begin{bmatrix} 1 & \dfrac { 1 }{ 2 } & 2 & 0 \\ 0 & \dfrac { 5 }{ 2 } & -2 & 0 \end{bmatrix}_{R1->R_1-\frac12R_2}$$ $$=\begin{bmatrix} 1 & \dfrac { 1 }{ 2 } & 2 & 0 \\ 0 & 1 & -\dfrac { 4 }{ 5 } & 0 \end{bmatrix}_{R1->R_1-\frac{R_2}{2}}$$ $$=\begin{bmatrix} 1 & 0 & \dfrac { 12 }{ 5 } & 0 \\ 0 & 1 & -\dfrac { 4 }{ 5 } & 0 \end{bmatrix}$$
$$x_1+\dfrac{12}{5}x_3=0$$ $$x_2-\dfrac45x_3=0$$ $$\mbox{Let $x_3=k$ be any arbitrary constant}$$ $$x_1=-\dfrac{12}{5}k\mbox{ and }x_2=\frac45k$$ $$\mbox{Therefor, the orthogonal complement or the basis}=\begin{bmatrix} -\dfrac { 12 }{ 5 } \\ \dfrac { 4 }{ 5 } \\ 1 \end{bmatrix}$$
$\endgroup$ 10 $\begingroup$The orthogonal complement is the set of all vectors whose dot product with any vector in your subspace is 0. It's a fact that this is a subspace and it will also be complementary to your original subspace.
In this case that means it will be one dimensional. So all you need to do is find a (nonzero) vector orthogonal to [1,3,0] and [2,1,4], which I trust you know how to do, and then you can describe the orthogonal complement using this.
$\endgroup$ $\begingroup$Take $(a,b,c)$ in the orthogonal complement. Then, since any element in the orthogonal complement must be orthogonal to $W=\langle(1,3,0)(2,1,4)\rangle$, you get this system:
$$(a,b,c) \cdot (1,3,0)= a+3b = 0$$ $$(a,b,c) \cdot (2,1,4)= 2a+b+4c = 0$$
One can see that $(-12,4,5)$ is a solution of the above system. Just take $c=1$ and solve for the remaining unknowns.
Of course, any $\vec{v}=\lambda(-12,4,5)$ for $\lambda \in \mathbb{R}$ is also a solution to that system. Since we are in $\mathbb{R}^3$ and $\dim W = 2$, we know that the dimension of the orthogonal complement must be $1$ and hence we have fully determined the orthogonal complement, namely:
$$W^{\perp}=\langle (-12,4,5)\rangle$$
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