Q.Let $\hat a,\hat b,\hat c$ be unit vectors such that $\hat a\cdot\hat b=\hat a\cdot\hat c=0$ and the angle between $\hat b$ and $\hat c$ be $\pi/6$. Prove that $\hat a=\pm2(\hat b\times\hat c)$.

The vector (cross) product combines two vectors to produce a third vector: $\vec A\times\vec B=(AB\sin\theta)\hat n$, where $\theta$ is the angle between $\vec A$ and $\vec B$, and $\hat n$ is perpendicular to the plane containing both, its sense fixed by the right-hand (screw) rule — curl the fingers from $\vec A$ toward $\vec B$ through the smaller angle; the thumb gives $\hat n$. It is not commutative: $\vec A\times\vec B=-(\vec B\times\vec A)$. It is maximum (magnitude $AB$) when the vectors are perpendicular ($\theta=90°$) and zero when they are parallel or anti-parallel ($\theta=0°$ or $180°$); in particular $\vec A\times\vec A=0$ always. For orthogonal unit vectors, $\hat i\times\hat j=\hat k$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat i=\hat j$ (and the reverse orderings give the negatives). In component form, using the determinant recipe,

$\vec A\times\vec B=(A_yB_z-A_zB_y)\hat i+(A_zB_x-A_xB_z)\hat j+(A_xB_y-A_yB_x)\hat k$ …