Q.If $\bar a\cdot\bar b=\sqrt3$ and $\bar a\times\bar b=2\hat i+\hat j+2\hat k$, find the angle between $\bar a$ and $\bar b$.

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,

Geometrically, if $\vec A$ and $\vec B$ are adjacent sides of a parallelogram, its area is $|\vec A\times\vec B|$, and a triangle with sides $\vec A,\vec B$ has area $\tfrac12|\vec A\times\vec B|$. Physically, every rotational quantity is built from a cross product: torque $\vec\tau=\vec r\times\vec F$, angular momentum $\vec L=\vec r\times\vec p$, and linear velocity from angular velocity, $\vec v=\vec\omega\times\vec r$.