6.2. THE GEOMETRIC SIGNIFICANCE OF THE DOT PRODUCT 81

Proof: Suppose 6.14 and F || = αD. Taking the dot product of both sides with D andusing F⊥ ·D = 0, this yields

F ·D = α |D|2

which requires α =F ·D/ |D|2 . Thus there can be no more than one vector F ||. It followsF⊥ must equal F −F ||. This verifies there can be no more than one choice for both F ||and F⊥.

Now letF || ≡

F ·D|D|2

D

and letF⊥ = F −F || = F−F ·D

|D|2D

Then F || = α D where α = F ·D|D|2

. It only remains to verify F⊥ ·D = 0. But

F⊥ ·D = F ·D−F ·D|D|2

D ·D = F ·D−F ·D = 0.

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Example 6.2.9 Let F = 2i+7j−3k Newtons. Find the work done by this force in movingfrom the point (1,2,3) to the point (−9,−3,4) along the straight line segment joining thesepoints where distances are measured in meters.

According to the definition, this work is

(2 i+7j−3k) · (−10i−5j+k) =−20+(−35)+(−3) =−58 Newton meters.

Note that if the force had been given in pounds and the distance had been given in feet,the units on the work would have been foot pounds. In general, work has units equal tounits of a force times units of a length. Instead of writing Newton meter, people write joulebecause a joule is by definition a Newton meter. That word is pronounced “jewel” and it isthe unit of work in the metric system of units. Also be sure you observe that the work doneby the force can be negative as in the above example. In fact, work can be either positive,negative, or zero. You just have to do the computations to find out.

Example 6.2.10 Find proju (v) if u= 2i+3j−4k and v = i−2j+k.

From the above discussion in Theorem 6.2.8, this is just

14+9+16

(i−2j+k) · (2i+3j−4k)(2i+3j−4k)

=−829

(2i+3j−4k) =−1629

i− 2429

j+3229

k.

Example 6.2.11 Suppose a, and b are vectors and b⊥ = b− proja (b) . What is the mag-nitude of b⊥ in terms of the included angle?