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?
Best Answer
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 $\left\langle -\frac{4}{3}, -\frac{2}{3}, 0, 1\right\rangle$