If `veca, vecb and vecc` are the position vectors of the vertices A,B and C. respectively , of `triangleABC`. Prove that the perpendicualar distance of the vertex A from the base BC of the triangle ABC is `(|vecaxxvecb+vecbxxvecc+veccxxveca|)/(|vecc-vecb|)`

Explanation

`|vec(BC)xxvec(BA)|=|vecaxxvecb+vecb+vecbxxvecc+veccxxveca|`
`|vec(BC)||vec(BA)|sin B=|vecaxxvecbxxvecbxxvecc+veccxxveca|`
`|vecc-vecb| (ABsinB)=|vecaxxvecb+vecbxxvecc+veccxxveca|`
Therefore, the length of perpendicualr from A on BC is
`AL=ABsinB=(|vecaxxvecb+vecbxxvecc+veccxxveca|)/(|vecb-vecc|)`