Find the unit vector normal to the plane determined by vec…

Find the unit vector normal to the plane determined by $\vec a=\{1,1,0\}$ and $\vec b=\{1,-1,1\}$.

Explanation of Method

The unit vector normal to the plane of two vectors.

\[ \vec a\times\vec b=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&0\\1&-1&1\end{vmatrix}=\vec i(1-0)-\vec j(1-0)+\vec k(-1-1)=\{1,-1,-2\}. \]

Check: $\{1,-1,-2\}\cdot\vec a=0$ and $\{1,-1,-2\}\cdot\vec b=1+1-2=0$ $\checkmark$

\[ |\vec a\times\vec b|=\sqrt{1+1+4}=\sqrt6. \]

\[\vec n_0=\pm\frac{1}{\sqrt6}\{1,-1,-2\}=\pm\left\{\tfrac{\sqrt6}{6},-\tfrac{\sqrt6}{6},-\tfrac{\sqrt6}{3}\right\}\]

Both signs are unit normals. The $+$ sign gives the one for which $\vec a,\vec b,\vec n_0$ form a right-handed triple.

$O$ $\vec a$ $\vec b$ $\vec n_0$

$\vec n_0$ is perpendicular to the shaded plane of $\vec a$ and $\vec b$; $-\vec n_0$ is the other unit normal.

\[\boxed{\,\vec n_0=\pm\frac{1}{\sqrt6}\{1,-1,-2\}=\pm\left\{\tfrac{\sqrt6}{6},-\tfrac{\sqrt6}{6},-\tfrac{\sqrt6}{3}\right\}\,}\]