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DOT PRODUCT EXAMPLES


EXAMPLE 1
Find the dot product, A⋅B, where:
(1) A  =   < 2, 5, 7 >         B  =   <  − 3, 1, 1 > 
This is a commonly used vector notation with the form  < i, j, k > . The solution to this problem begins by multiplying the two i components together, then computing the product of the two j and finally the two k components. These three products are then added to yield
(2) A⋅B  =  2⋅( − 3) + 5⋅1 + 7⋅1        =   − 6 + 5 + 7        =  6
It is important to remember that the unit vectors do not have to be in the same order, so the reader is cautioned to examine each vector carefully to ensure that the correct coefficients are being multiplied together.

EXAMPLE 2
Find the angle between the vectors A and B if
(3) A  =  6 i − 2 j − 3 k =  < 6,  − 2,  − 3 >         B  =  1 i + 1 j + 1 k =  < 1, 1, 1 > 
The first step in solving this problem is recall the angular relationship that defines the dot product:
(4) A⋅B = |A||B|cosθ
Rearranging so that the answer, θ, is on the left hand side of the equation, the equation becomes:
(5) cosθ = (A⋅B)/(|A||B|)
The magnitude of each vector can be found by squaring the coefficient of each unit vector, then taking the square root of their sums. Substituting the values from Eqn. ↓ yields
(6) cosθ  =  ((6⋅1) + ( − 2⋅1) + ( − 3⋅1))/(√((6)2 + ( − 2)2 + ( − 3)2)⋅√((1)2 + (1)2 + (1)2))        =  (1)/(√(49)⋅√(3)) = (1)/(147)
and
(7) θ  =  cos − 1⎛⎝(1)/(147)⎞⎠        =  81.61

EXAMPLE 3
Are the vectors A =  < 2, 2,  − 2 >  and B =  < 4,  − 2, 2 >  orthogonal with one another?
Two vectors are orthogonal if their dot product is equal to zero, so the dot product simply must be computed, and the value examined:
(8) A⋅B  =  2⋅4 + 2⋅( − 2) + ( − 2)⋅2        =  8 − 4 − 4        =  0
Since the answer is zero, the two vectors are indeed orthogonal or normal to one another.