Obtain scalar product in terms of Cartesian component of vectors .

Explanation

`vec(A) and vec(B)` is written in Cartesian component as follow :
`vec(A) = A_(x) hat(i) +A_(y) hat(j) +A_(z)hat(k)`
`vec(A) = A_(x) hat(i) +A_(y)hat(j) +A_(z)hat(k)`
`vec(B) = B_(x) hat(i) = B_(y) hat(j) +B_(z)hat(k)`
` :. vec(A) .vec(B) = (A_(x)hat(i) +A_(y)hat(j)+A_(z)hat(k)).(B_(x)hat(i) +B_(y)hat(j) +B_(z)hat(k))`
`A_(x)B_(x)(hat(i).hat(i))+A_(x) B_(y)(hat(i).hat(j))+A_(z)B_(z)(hat(i).hat(k))`
` + A_(y)B_(x) (hat(j).hat(i)) +A_(y)B_(z) (hat(j).hat(k))`
`+A_(z)B_(x)(hat(k).hat(i))+A_(z)B_(y)(hat(k).hat(j))+A_(z)B_(z)(hat(k).hat(k))`
In this equation `hat(i) . hat(i) = hat(j) .hat(j) = hat(k) .hat(k) = 1 ` and `hat(i).hat(j)= hat(j).hat(i) = 0 , hat(j) . hat(k) = hat(k) . hat(j) = 0 ` and `hat(k).hat(i) = hat(i) .hat(k) = 0 `
So , `vec(A) . vec(B) =A_(x)B_(x) +A_(y)B_(y) +A_(z)B_(z)`