Vector perpendicular to three other vectors Find a vector $$\vec{w}$$ such that it is perpendicular to the following vectors:
$$
\vec{v_1} = [-1,2,-1,0]^{\,T}  
$$
$$
\vec{v_2} = [-3,9,-3,2]^{\,T}
$$
$$
\vec{v_3} = [-1,-10,2,-8]^{\,T}
$$
Now I know that in order for two vectors to be perpendicular, the dot product must equal zero. 
Can I take the three vectors, augment them into a 4X3 matrix and then some how get it into the form Ax = b? 
 A: Note that a vector $\vec v$ is orthogonal to your given vectors if and only if $A\vec v=\vec 0$ where
$$
A=
\left[\begin{array}{rrrr}
-1 & 2 & -1 & 0 \\
-3 & 9 & -3 & 2 \\
-1 & -10 & 2 & -8
\end{array}\right]
$$
Thus we wish to compute a basis for the null space of $A$. Row-reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rrrr}
1 & 0 & 0 & \frac{4}{3} \\
0 & 1 & 0 & \frac{2}{3} \\
0 & 0 & 1 & 0
\end{array}\right]
$$
Hence a $\vec v$ is orthogonal to your given vectors if and only if
$$
\vec v = 
\begin{bmatrix}v_1\\ v_1\\ v_3\\ v_4\end{bmatrix}=
\begin{bmatrix}-\frac{4}{3}\,v_4\\ -\frac{2}{3}\,v_4\\0\\ v_4\end{bmatrix}=
\begin{bmatrix}-\frac{4}{3}\\ -\frac{2}{3}\\0\\ 1\end{bmatrix}v_4
$$
That is, the vectors orthogonal to your given vectors are exactly the scalar multiples of $\left\langle -\frac{4}{3}, -\frac{2}{3}, 0, 1\right\rangle$
A: Or take their cross product :
$$v_1\times v_2\times v_3=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\-3&9&-3&2\\-1&-10&2&-8\end{vmatrix}=$$$${}$$
$$=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\0&3&0&2\\0&-12&3&-8\end{vmatrix}=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\0&3&0&2\\0&0&3&0\end{vmatrix}=-3\begin{vmatrix}i&j&w\\-1&2&0\\0&3&2\end{vmatrix}=\left(-12,\,-6,\,0,\,9\right)=:u$$$${}$$
Now check $\;u^t\;$ indeed perpendicular to all your three vectors.
