Our first application will be to the concept of work. The physical concept of work does not in any way correspond to the notion of work employed in ordinary conversation. For example, if you were to slide a 150 pound weight off a table which is three feet high and shuffle along the floor for 50 yards, sweating profusely and exerting all your strength to keep the weight from falling on your feet, keeping the height always three feet and then deposit this weight on another three foot high table, the physical concept of work would indicate that the force exerted by your arms did no work during this project even though the muscles in your hands and arms would likely be very tired. The reason for such an unusual definition is that even though your arms exerted considerable force on the weight, enough to keep it from falling, the direction of motion was at right angles to the force they exerted. The only part of a force which does work in the sense of physics is the component of the force in the direction of motion (This is made more precise below.). The work is defined to be the magnitude of the component of this force times the distance over which it acts in the case where this component of force points in the direction of motion and
In this picture the force, F is applied to an object which moves on the straight line from p_{1} to p_{2}. There are two vectors shown, F_{} and F_{⊥} and the picture is intended to indicate that when you add these two vectors you get F while F_{} acts in the direction of motion and F_{⊥} acts perpendicular to the direction of motion. Only F_{} contributes to the work done by F on the object as it moves from p_{1} to p_{2}. F_{} is called the component of the force in the direction of motion. From trigonometry, you see the magnitude of F_{} should equal

If the included angle had been obtuse, then the work done by the force, F on the object would have been negative because in this case, the force tends to impede the motion from p_{1} to p_{2} but in this case, cosθ would also be negative and so it is still the case that the work done would be given by the above formula. Thus from the geometric description of the dot product given above, the work equals

This explains the following definition.
Definition 3.2.7 Let F be a force acting on an object which moves from the point p_{1} to the point p_{2}. Then the work done on the object by the given force equals F⋅
The concept of writing a given vector F in terms of two vectors, one which is parallel to a given vector D and the other which is perpendicular can also be explained with no reliance on trigonometry, completely in terms of the algebraic properties of the dot product. As before, this is mathematically more significant than any approach involving geometry or trigonometry because it extends to more interesting situations. This is done next.
Theorem 3.2.8 Let F and D be nonzero vectors. Then there exist unique vectors F_{} and F_{⊥} such that
 (3.14) 
where F_{} is a scalar multiple of D, also referred to as

and F_{⊥}⋅ D = 0. The vector proj _{D}
Proof: Suppose 3.14 and F_{} = αD. Taking the dot product of both sides with D and using F_{⊥}⋅ D = 0, this yields

which requires α = F ⋅ D∕
Now let

and let

Then F_{} = α D where α =

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Example 3.2.9 Let F = 2i+7j − 3k Newtons. Find the work done by this force in moving from the point
According to the definition, this work is

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 to units of a force times units of a length. Instead of writing Newton meter, people write joule because a joule is by definition a Newton meter. That word is pronounced “jewel” and it is the unit of work in the metric system of units. Also be sure you observe that the work done by 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 3.2.10 Find proj _{u}
From the above discussion in Theorem 3.2.8, this is just
Example 3.2.11 Suppose a, and b are vectors and b_{⊥} = b − proj _{a}
