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

Apr 15, 2017

$A . B = | A | | B | \cos \alpha$
$\alpha$ is the angle between vector A and vector B .

Explanation:

$A . B = \left({A}_{1} i + {A}_{2} j + {A}_{3} k\right) . \left({B}_{1} i + {B}_{2} j + {B}_{3} k\right)$
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 .
$A . B = \left({A}_{1} i . {B}_{1} i\right) + \left({A}_{2} j . {B}_{2} j\right) + \left({A}_{3} k . {B}_{3} k\right)$
dot product two vectors with same unit vector will just be product of their magnitudes as angle between them is zero .
THEREFORE ,
$A . B = {A}_{1} {B}_{1} + {A}_{2} {B}_{2} + {A}_{3} {B}_{3}$

Apr 15, 2017

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: $\vec{A} \cdot \vec{B} = \left\mid A \right\mid \left\mid B \right\mid \cos \alpha$.

We are using the Cartesian $\left\langle \hat{i} , \hat{j} , \hat{k}\right\rangle$ basis which can be written as $\left\langle {\hat{x}}_{1} , {\hat{x}}_{2} , {\hat{x}}_{3}\right\rangle$ to allow the following notation.

From the geometric definition:

$\vec{A} \setminus \cdot \vec{B} = {\sum}_{i = 1}^{3} {A}_{i} {\hat{x}}_{i} \setminus \cdot {\sum}_{j = 1}^{3} {B}_{j} {\hat{x}}_{j} q \quad \triangle$

Because of the orthogonality we have:

${\left({\hat{x}}_{i} \setminus {\hat{x}}_{j}\right)}_{i = j} = 1$

${\left({\hat{x}}_{i} \setminus {\hat{x}}_{j}\right)}_{i \ne j} = 0$

So $\triangle$ simplifies:

$\vec{A} \setminus \cdot \vec{B} = {A}_{1} {B}_{1} + {A}_{2} {B}_{2} + {A}_{3} {B}_{3}$