If A= A1i + A2j + A3k and B= B1i + B2j + B3k. Prove that A.B= A1B1+A2B2+A3B3?

`A.B = ( A_1 i + A_2 j + A_3 k ).( B_1 i +B_2 j +B_3 k)`
NOW , i and j unit vectors are perpendicular to each other and so are the vectors j and k & i and k .
thus , dot product of such vectors is zero .
`A.B = ( A_1 i . B_1 i ) + ( A_2 j . B_2 j ) + ( A_3 k . B_3 k )`
dot product two vectors with same unit vector will just be product of their magnitudes as angle between them is zero .
THEREFORE ,
`A.B = A_1B_1 + A_2B_2 + A_3B_3`

What you've presented is the algebraic definition of the dot product so I guess you're trying to connect that to the geometric definition which is: `vec A cdot vec B = abs A abs B cos alpha`.

We are using the Cartesian `langlehat i, hat j,hat k rangle` basis which can be written as `langlehat x_1, hat x_2 ,hat x_3rangle` to allow the following notation.

From the geometric definition:

`vec A \cdot vec B =sum_(i = 1)^3 A_i hat x_i \cdot sum_(j = 1)^3 B_j hat x_j qquad triangle`

Because of the orthogonality we have:

`(hat x_i \ hat x_j)_(i = j) = 1`

`(hat x_i \ hat x_j)_(i ne j) = 0`

So `triangle` simplifies:

`vec A \cdot vec B = A_1B_1 + A_2 B_2 + A_3 B_3`