A unit vector $\vec{a}$ in the plane $\vec{b}=2\hat{i}+\hat{j}$ & $\vec{c}=\hat{i}-\hat{j}+\hat{k}$ of is such that $\vec{a}$^$\vec{b}$ = $\vec{a}$^$\vec{d}$ where $\vec{d}=\hat{j}+2\hat{k}$ is:
What does $\vec{a}$^$\vec{b}$ mean?? Is it a misprint or a new operator??
For general vectors $v, w$ in a vector space $V$, the notation $v\land w$ usually denotes their exterior product, which can be thought of as a directed parallelogram (spanned by $v$ and $w$), where two parallelograms are regarded equal if they lie on the same plane, they have the same area and direction. (See 'exterior algebra' on wikipedia.)
In $\Bbb R^3$ this notion essentially coincides with the cross product.
More specifically, the space of these parallelograms of $\Bbb R^3$ with above equality constrain is isomorphic to $\Bbb R^3$ via the mapping $v\land w\mapsto v\times w$.
The advantage of the exterior product is that it can be defined for any vector space, whereas the cross product lives only in $3$ dimensions, though the result of the exterior product lies in another space.