Matrix Theory: Orthogonal Matrix Consider a vector   
$$
  \begin{array}{l}
    a=(1,-3,0,1), \\
    b=(1,2,0,-1), \\ 
    c=(0,0,1,0)
  \end{array}
$$
Question: 
(a) A non-zero vector $u$ orthogonal to all three of $a, b, c$ (there are many, just find one). 
(b) Describe the set of all vectors with these properties using a parameter or parameters.
Can anyone tell me how to solve this?
 A: (a) $$u=(u_1,u_2,u_3,u_4)$$
So that $u$ is orthogonal to $a$: $u \cdot a=0  \Rightarrow u_1-3u_2+u_4=0 \ \ \ (1)$
So that $u$ is orthogonal to $b$: $u \cdot b=0  \Rightarrow u_1+2u_2-u_4=0 \ \ \ (2)$
So that $u$ is orthogonal to $c$: $u \cdot b=0  \Rightarrow u_3=0$
$$(1)+(2) \Rightarrow 2u_1=u_2$$
For $u_1=1$ we have $u_2=2$.
$(1) \Rightarrow 1-6+u_4=0 \Rightarrow u_4=5$.
Therefore, a non-zero vector $u$ orthogonal to all three of $a, b, c$ is $(1,2,0,5)$.
EDIT:
(b)
From the above relations we have that:
$$2u_1=u_2$$
$$u_3=0$$
$$\text{ Replacing } 2u_1=u_2 \text{ at the relation } (2) \text{ we get: } u_1+4u_1-u_4=0 \Rightarrow u_4=5u_1$$
Therefore, $$u=(u_1,u_2,u_3,u_4)=(u_1,2u_1, 0, 5u_1)=u_1(1,2,0,5)$$ 
So, the set of all vectors with these properties is: $$\{u_1(1,2,0,5):u_1 \in \mathbb{R} ^*\}$$
A: Maybe you already realize this since you've used the tag (matrices), but you can solve this problem by computing the kernel of the matrix
$$
\begin{pmatrix}
1 & -3 & 0 & 1\\
1 & 2 & 0 & -1\\
0 & 0 & 1 & 0
\end{pmatrix} \, .
$$
According to Wolfram Alpha, this matrix has RREF
$$
\begin{pmatrix}
1 & 0 & 0 & -1/5\\
0 & 1 & 0 & -2/5\\
0 & 0 & 1 & 0
\end{pmatrix}
$$
which should allow you to read the answer off.
