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  4. The dot product

2.3 The dot product  (Page 6/16)

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q · p = ⟨ 9 5 , − 27 5 ⟩ · ⟨ 21 5 , 7 5 ⟩ = 9 ( 21 ) 25 + −27 ( 7 ) 25 = 189 25 − 189 25 = 0.

Therefore, q and p are orthogonal.

Resolving vectors into components

Express v = ⟨ 8 , −3 , −3 ⟩ as a sum of orthogonal vectors such that one of the vectors has the same direction as u = ⟨ 2 , 3 , 2 ⟩ .

Let p represent the projection of v onto u :

p = proj u v = u · v ‖ u ‖ 2 u = ⟨ 2 , 3 , 2 ⟩ · ⟨ 8 , −3 , −3 ⟩ ‖ ⟨ 2 , 3 , 2 ⟩ ‖ 2 ⟨ 2 , 3 , 2 ⟩ = 16 − 9 − 6 2 2 + 3 2 + 2 2 ⟨ 2 , 3 , 2 ⟩ = 1 17 ⟨ 2 , 3 , 2 ⟩ = ⟨ 2 17 , 3 17 , 2 17 ⟩ .

Then,

q = v − p = ⟨ 8 , −3 , −3 ⟩ − ⟨ 2 17 , 3 17 , 2 17 ⟩ = ⟨ 134 17 , − 54 17 , − 53 17 ⟩ .

To check our work, we can use the dot product to verify that p and q are orthogonal vectors:

p · q = ⟨ 2 17 , 3 17 , 2 17 ⟩ · ⟨ 134 17 , − 54 17 , − 53 17 ⟩ = 268 17 − 162 17 − 106 17 = 0 .

Then,

v = p + q = ⟨ 2 17 , 3 17 , 2 17 ⟩ + ⟨ 134 17 , − 54 17 , − 53 17 ⟩ .
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Express v = 5 i − j as a sum of orthogonal vectors such that one of the vectors has the same direction as u = 4 i + 2 j .

v = p + q , where p = 18 5 i + 9 5 j and q = 7 5 i − 14 5 j

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Scalar projection of velocity

A container ship leaves port traveling 15 ° north of east. Its engine generates a speed of 20 knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of 2 knots. Considering both the engine and the current, how fast is the ship moving in the direction 15 ° north of east? Round the answer to two decimal places.

This figure is an image of a ship. The ship is at the origin of two perpendicular axes. The horizontal axis is labeled “east.” The second axis is vertical and labeled “north.” From the ship there are two vectors. The first is labeled “v” and has an angle of 15 degrees between the East axis and the vector v. The second vector is labeled “w” and has an angle of 45 degrees between the East axis and the vector w.

Let v be the velocity vector generated by the engine, and let w be the velocity vector of the current. We already know ‖ v ‖ = 20 along the desired route. We just need to add in the scalar projection of w onto v . We get

comp v w = v · w ‖ v ‖ = ‖ v ‖ ‖ w ‖ cos ( 30 ° ) ‖ v ‖ = ‖ w ‖ cos ( 30 ° ) = 2 3 2 = 3 ≈ 1.73 knots .

The ship is moving at 21.73 knots in the direction 15 ° north of east.

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Repeat the previous example, but assume the ocean current is moving southeast instead of northeast, as shown in the following figure.

This figure is an image of a ship. The ship is at the origin of two perpendicular axes. The horizontal axis is labeled “east.” The second axis is vertical and labeled “north.” From the ship there are two vectors. The first is labeled “v” and has an angle of 15 degrees between the East axis and the vector v. The second vector is labeled “w” and has an angle of 45 degrees between the East axis and the vector w. Vector w is below the East axis in the fourth quadrant.

21 knots

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Work

Now that we understand dot products, we can see how to apply them to real-life situations. The most common application of the dot product of two vectors is in the calculation of work.

From physics, we know that work is done when an object is moved by a force. When the force is constant and applied in the same direction the object moves, then we define the work done as the product of the force and the distance the object travels: W = F d . We saw several examples of this type in earlier chapters. Now imagine the direction of the force is different from the direction of motion, as with the example of a child pulling a wagon. To find the work done, we need to multiply the component of the force that acts in the direction of the motion by the magnitude of the displacement. The dot product allows us to do just that. If we represent an applied force by a vector F and the displacement of an object by a vector s , then the work done by the force is the dot product of F and s .

Definition

When a constant force is applied to an object so the object moves in a straight line from point P to point Q , the work W done by the force F , acting at an angle θ from the line of motion, is given by

W = F · P Q → = ‖ F ‖ ‖ P Q → ‖ cos θ .

Let’s revisit the problem of the child’s wagon introduced earlier. Suppose a child is pulling a wagon with a force having a magnitude of 8 lb on the handle at an angle of 55°. If the child pulls the wagon 50 ft, find the work done by the force ( [link] ).
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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