Find the vector vec x perpendicular to vec a=\3,2,-1\ , vec…

Find the vector $\vec x$ perpendicular to $\vec a=\{3,2,-1\}$, $\vec b=\{2,-1,3\}$ and $\vec c=\{1,1,-1\}$. Is there more than one such vector?

Step 1: System

\[ \begin{aligned} 3x_1+2x_2-x_3&=0 &&(\mathrm I)\\ 2x_1-x_2+3x_3&=0 &&(\mathrm{II})\\ x_1+x_2-x_3&=0 &&(\mathrm{III}) \end{aligned} \]

Step 2: Solving

From (III), $x_1=x_3-x_2$. Then (I) gives $-x_2+2x_3=0$, i.e. $x_2=2x_3$. (II) gives $5x_3-3x_2=0$, i.e. $-x_3=0$. Hence

\[ x_3=0,\quad x_2=0,\quad x_1=0. \]

Step 3: Why only the zero vector

\[ \det\begin{pmatrix}3&2&-1\\2&-1&3\\1&1&-1\end{pmatrix}=3(1-3)-2(-2-3)-(2+1)=1\neq0. \]

The three vectors are not coplanar, so they span the whole space. A vector perpendicular to all three would be perpendicular to itself: $\vec x\cdot\vec x=0$, so $\vec x=\vec 0$.

\[\vec x=\vec 0\ \ \text{is the only solution; no non-zero vector exists}\]

Contrast. For three coplanar vectors (determinant $0$), there would be infinitely many solutions: every multiple of the normal of their plane.