Answer
\[
1=\vec{u} \cdot(\vec{w} \times \vec{v})
\]
Work Step by Step
To evaluate $ (\vec{w} \times \vec{v})\cdot\vec{u}$, we can use the following formula.
\[
(\vec{w} \times \vec{v})\cdot\vec{u} =\left|\begin{array}{lll}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\]
It's given that
\[
\begin{array}{l}
\vec{w}=\hat{i}+\hat{j}+\hat{k}=\langle1,1,1\rangle , \quad \vec{v}=\hat{i}+\hat{j}=\langle 1,1,0\rangle \text { and } \\
\vec{u}=\hat{i}=\langle 1,0,0\rangle
\end{array}
\]
And then,
\[
\begin{aligned}
(\vec{w} \times \vec{v}) \cdot\vec{u}&=\left|\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{array}\right| \\
&=1\left|\begin{array}{cc}
1 & 0 \\
1 & 1
\end{array}\right|-0\left|\begin{array}{cc}
1 & 0 \\
1 & 1
\end{array}\right|+0\left|\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array}\right| \\
&=1(1-0)=1
\end{aligned}
\]
We found that:
\[
1=(\vec{w} \times \vec{v})\cdot\vec{u}
\]
