# Consider the plane, ccP in RR^3 by the vector equation x(s,t)=(1,−1,2)+s(1,0,1)+t(1,−1,0);s,t in RR Compute a unit normal vector, n, to this plane.

Consider the plane, 𝒫 in ${ℝ}^{3}$ by the vector equation
$x\left(s,t\right)=\left(1,-1,2\right)+s\left(1,0,1\right)+t\left(1,-1,0\right);s,t\in ℝ$
Compute a unit normal vector, n, to this plane.
My attempt is the third normal vector is $\left(1,\frac{2s}{t}+1,1\right)$ and the unit normal vector I got is
$\frac{1}{\sqrt{3+\frac{4{s}^{2}}{{t}^{2}}+\frac{4s}{t}}}\left(1,\frac{2s}{t}+1,1\right)$
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HINT
$\mathbf{\text{n}}=\frac{\left(1,0,1\right)×\left(1,-1,0\right)}{‖\left(1,0,1\right)×\left(1,-1,0\right)‖}$
EDIT
Since $\left(1,0,1\right)=\mathbf{\text{i}}+\mathbf{\text{k}}$ and $\left(1,-1,0\right)=\mathbf{\text{i}}-\mathbf{\text{j}}$, one has that
$\begin{array}{rl}\left(1,0,1\right)×\left(1,-1,0\right)& =\left(\mathbf{\text{i}}+\mathbf{\text{k}}\right)×\left(\mathbf{\text{i}}-\mathbf{\text{j}}\right)=-\mathbf{\text{k}}+\mathbf{\text{j}}+\mathbf{\text{i}}=\left(1,1,-1\right)\end{array}$