Scalar Product

of Two Vectors

By definition, the scalar product, also known as the "dot product",

of two vectors A and B is

A B = A B cos

A B = A B cos

This "dot product" is commutative,

A B = B A

This "dot product" is distributative,

A (B + C) = A B + A C

A = A_x i + A_y j + A_zk

B = B_x i + B_y j + B_zk

A B = (A_x i + A_y j + A_zk) (B_x i + B_y j + B_zk)

= A_x B_x( i i) + A_x B_y( i j) + A_x B_z( i k) +

+ A_y B_x( j i) + A_y B_y( j j) + A_y B_z( j k) +

+ A_z B_x( k i) + A_z B_y( k j) + A_z B_z( k k)

What are these dot products of the unit vectors, i, j, and k?

From our initial definition of the scalar product, A B = A B cos , we have

i i = 1 1 cos 0 = 1	i j = 1 1 cos 90o = 0	i k = 1 1 cos 90o = 0
j i = 1 1 cos 90o = 0	j j = 1 1 cos 0 = 1	j k = 1 1 cos 90o = 0
k i = 1 1 cos 90o = 0	k j = 1 1 cos 90o = 0	k k = 1 1 cos 0 = 1

i i = 1 1 cos 0 = 1	i j = 1 1 cos 90o = 0	i k = 1 1 cos 90o = 0
j i = 1 1 cos 90o = 0	j j = 1 1 cos 0 = 1	j k = 1 1 cos 90o = 0
k i = 1 1 cos 90o = 0	k j = 1 1 cos 90o = 0	k k = 1 1 cos 0 = 1

That means

A B = (A_x i + A_y j + A_zk) (B_x i + B_y j + B_zk)

= A_x B_x( i i) + A_x B_y( i j) + A_x B_z( i k) +

+ A_y B_x( j i) + A_y B_y( j j) + A_y B_z( j k) +

+ A_z B_x( k i) + A_z B_y( k j) + A_z B_z( k k)

A B = A_x B_x( 1) + A_x B_y( 0) + A_x B_z( 0) +

+ A_y B_x( 0) + A_y B_y( 1) + A_y B_z( 0) +

+ A_z B_x( 0) + A_z B_y( 0) + A_z B_z( 1)

= A_x B_x + A_y B_y + A_z B_z

That is,

A B = A_x B_x + A_y B_y + A_z B_z

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The definition of the scalar product,

A B = A B cos

means that work is the scalar product of the Force and the displacement,

W = F s

Sometimes it will be easier to evaluate this as

W = F s = F s cos

and sometimes it will be easier to evaluate this as

W = F s = F_x s_x + F_y s_y + F_z s_z

W = F s = F_x x + F_y y + F_z z

Constant Force			Varying Force
Return to ToC, Ch7, Work and Energy

(c) Doug Davis, 2001; all rights reserved