How to find value of $\vec{u}\cdot(\vec{v}×\vec{w})$? Let $u$ and $v$ be unit vectors and $\vec{w}$ is a vector(not necessarily unit vector)  such $\vec{u}×\vec{v}+\vec{u}=\vec{w}$ and $\vec{w}×u=v$,then how to find value of $\vec{u}\cdot(\vec{v}×\vec{w})$ ?
What would be the best and shortest approach to this problem? Should I assume certain angles between the vectors say $\alpha,\beta,\gamma$ and try subtituting in the equations? But that is too long. Help needed.
 A: Hint: the triple product is invariant under cyclic permutations of the three vectors.
More details as requested: since the triple product is invariant under cyclic permutations,
$$u\cdot (v \times w) = v \cdot (w \times u) = v \cdot v = 1.$$
Notice that the other condition $u \times v + u = w$ is unneeded here.
A: Since $\vec{w}=\vec{u}\times\vec{v}+\vec{u}$ it follows
\begin{align}
\vec{v}\times\vec{w}&=\vec{v}\times\left(\vec{u}\times\vec{v}\right)+\vec{v}\times\vec{u}\\
&=\left(\vec{v}\cdot\vec{v}\right)\vec{u}-\left(\vec{v}\cdot\vec{u}\right)\vec{v}+\vec{v}\times\vec{u}\qquad\quad\text{from the properties of the cross product}\\
&=\vec{u}-\left(\vec{v}\cdot\vec{u}\right)\vec{v}+\vec{v}\times\vec{u}\qquad\qquad\quad\;\;\,\text{since }\vec{v}\cdot\vec{v}=\left\|\vec{v}\right\|^2=1\\
&=\vec{u}+\vec{v}\times\vec{u}
\end{align}
Cause $\vec{w}\times\vec{u}=\vec{v}$ implies $\vec{v}$ is perpendicular to $\vec{u}$ and so $\vec{v}\cdot\vec{u}=0$, thus
\begin{align}
\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)&=\vec{u}\cdot\vec{u}+\vec{u}\cdot\left(\vec{v}\times\vec{u}\right)\\
&=\left\|\vec{u}\right\|^2+0\qquad\qquad\quad\text{cause }\vec{u}\times\vec{v}\,\text{ is perpendicular to }\vec{u}\\
&=1
\end{align}
A: $v×w=(w×u)×w=(w.w)u-(w.u)w$ using property of vector triple product.
Now $w.u=(u×v+u).u=0+1=1$
and,
$w.w=(u×v).(u×v)+u.u=1+1=2$
Hence,required result is $u.[(w.w)u-(w.u)w]=2.1-1.1=1$
A: $\vec{v}\times\vec{w}=\vec{v}\times(\vec{u}\times\vec{v})+\vec{v}\times\vec{u}=(\vec{v}\cdot\vec{v})\vec{u}-(\vec{v}\cdot\vec{u})\vec{v}+\vec{v}\times\vec{u}=\vec{u}+\vec{v}\times\vec{u},\;$ so
$\vec{u}\cdot(\vec{v}\times\vec{w})=\vec{u}\cdot(\vec{u}+\vec{v}\times\vec{u})=\vec{u}\cdot\vec{u}+\vec{u}\cdot(\vec{v}\times\vec{u})=1+0=1$
(using that $\vec{u}\cdot\vec{u}=1, \vec{v}\cdot\vec{v}=1$, and $\vec{u}\cdot\vec{v}=0$)
