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

2 Answers
Apr 15, 2017

Answer:

# A.B = |A||B|cosalpha #
#alpha# is the angle between vector A and vector B .

Explanation:

# 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 #

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 = 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#