Math 32A Discussion Session Week 2 Notes October 10 and 12, 2017 Since we didn t get a chance to discuss parametrized lines last week, we may spend some time discussing those before moving on to the dot product, which is discussed below. Last week s notes contain some remarks about parametrized lines. The Dot Product The first product we d like to define between vectors is one that measures the extent to which the vectors are pointing in the same direction. That is, if u and v are two vectors pointing in the same direction, their product should be positive, while if they point in opposite directions, their product should be negative. This product should also have a feature we don t see in products of numbers: it should be possible for a product of two vectors to be 0 even though neither of the vectors is zero. In particular, if u and v are orthogonal, or perpendicular, then their product should be 0, since they re pointing neither in the same nor the opposite direction from each other. It s possible to come up with a product strictly from these geometric properties, if a bit tedious. Instead we ll define the dot product algebraically, and then show that it has the properties we hoped for. The Dot Product. Suppose u and v are vectors with n components: Then the dot product of u with v is u = u 1, u 2,..., u n, v = v 1, v 2,..., v n. u v = u 1 v 1 + u 2 v 2 + + u n v n. Notice that the dot product of two vectors is a scalar, and also that u and v must have the same number of components in order for u v to be defined. For example, if u = 1, 2, 4, 2 and v = 2, 1, 0, 3, then u v = 1 2 + 2 1 + 4 0 + ( 2) 3 = 2. It s interesting to note that the dot product is a product of two vectors, but the result is not a vector. In particular, if u, v, and w are vectors, then u (v w) doesn t make any sense, because v w is a scalar. Before we verify that the dot product has the geometric properties we d hoped for, we ll point out some algebraic properties of the dot product. Theorem 1. Suppose u, v, and w are vectors with the same number of components as each other, λ is a scalar, and let 0 denote the zero vector. Then (i) 0 v = v 0 = 0. (ii) u v = v u. (iii) (λu) v = u (λv) = λ(u v). 1
Figure 1 (iv) u (v + w) = u v + u w and (u + v) w = u w + v w. Theorem 1 is telling us that the dot product shares several nice properties with the usual product of numbers. We ve already mentioned one that it does not share, however. Since, for example, 1, 0 0, 1 = 1 0 + 0 1 = 0, we see that the dot product admits zero divisors, meaning that u v can be 0 even when neither u nor v is 0. If we re comfortable with the algebra of the dot product, we can make our first connection with geometry. Recall that the length of a vector u is computed using the distance formula: u = u 2 1 + u2 2 + + u2 n. Then we can notice that u u = u 2 1 + u 2 2 + + u 2 n = u 2. So the dot product of a vector with itself is the square of the vector s length. This fits with our expectation that the product of two vectors pointing in the same direction be a positive number, since u 2 > 0 whenever u 0. In fact, we can use the observation that u u = u 2 to compute the angle θ between any two vectors u and v. Consider the vectors in Figure 1. We ve labeled the acute angle θ between them, but there s another angle 2π θ that also lies between them. Whenever we say let θ be the angle between two vectors, we assume that 0 θ π. That is, we always take θ to be the smaller of the two angles between the two vectors. From Figure 1 we can compute the length of the vector u v using the law of cosines: u v 2 = u 2 + v 2 2 u v cos θ. But using what we know about how the dot product is related to the length of a vector, we can also compute the length of u v using the dot product: u v 2 = (u v) (u v) = u u 2u v + v v = u 2 + v 2 2u v. So meaning that u 2 + v 2 2 u v cos θ = u 2 + v 2 2u v, u v = u v cos θ. 2
So we can compute the angle θ between u and v using the dot product: ( ) u v θ = arccos. u v Notice also that the dot product has the geometric properties we set out hoping for. If the vectors are pointing in the same direction, then θ = 0, so If the vectors are orthogonal, then θ = π 2, so u v = u v cos(0) = u v > 0. u v = u v cos(π/2) = 0. In fact, whenever the dot product between vectors u and v is positive, the angle between u and v is acute, meaning that u and v are pointing in the same general direction. If u v < 0, then the angle between u and v is obtuse. Example 1. The vectors in Figure 1 are u = 1, 2 and v = 3, 1. Compute θ. (Solution) First we compute the lengths of u and v: u = u u = 1 1 + 2 2 = 5, v = v v = 3 2 + 1 2 = 10. Since u v = 1 3 + 2 1 = 5, ( ) ( ) ( ) 5 5 1 θ = arccos = arccos = arccos 2 = π 5 10 10 4. So the angle θ between u and v is π/4. Example 2. Find the angle θ in the figure below. (Solution) Let P be the point (1 + 3, 1 + 3) and let Q be the point ( 1 + 3, 1 + 3). We ll let u be the vector whose baspoint is P and whose terminal point is the origin, and we ll let v have basepoint P and terminal point Q, so that u = 1 3, 1 3, and v = 2, 2. 3
The angle between u and v is θ, since these vectors agree with the sides of our triangle which are adjacent to θ. This means that cos θ = u v u v = ( 1 3) ( 2) + (1 3) 2 ( 1 3) 2 + (1 3) 2 = 4 = 1 ( 2) 2 + 2 2 8 8 2. Since the only angle θ satisfying 0 θ π and cos θ = 1/2 is π/3, we must have θ = π/3 in the figure. The dot product not only allows us to compute the angle between two vectors, it also allows us to compute the projection of one vector onto another. Consider the vectors u and v in Figure 2, which are the same as those in Figure 1. Since u = 1, 2, u has x-component 1 and y-component 2, but what is its v-component? That is, how much is u pointing in the v direction? One way to asses this is to project u onto v, producing a vector proj v u which points in the same direction as v and moves as much in the v-direction as does u. (Notice that if θ is obtuse, then proj v u actually points in the opposite direction from v.) So how do we go about finding proj v u? If 0 θ π/2, then proj v u points in the same direction as v, so proj v u = proj v u e v, where e v is the unit vector pointing in the direction of v. Since the triangle formed by u, proj v u, and the dashed line segment in Figure 2 is a right triangle, proj v u = u cos θ. Substituting this into our expression for proj v u, and using what we know about how the dot product relates to the angle between two vectors, we have proj v u = u cos θ v v = u u v u v Notice that proj v u can also be written in terms of the unit vector: We sometimes call u v v proj v u = u v v e v. the component of u along v. v v = u v v 2 v. Example 3. Compute proj v u, with u and v as in Figures 1 and 2. (Solution) Recall that u = 1, 2 and v = 3, 1, so is the projection of u onto v. proj v u = 1 3 + 2 1 3 2 + 1 2 3, 1 = 5 3, 1 = 3/2, 1/2 10 Example 4. ( 13.3, Exercise 81 of the textbook) Calculate the force (in newtons) required to push a 40-kg wagon up a 10, frictionless incline. 4
Figure 2 (Solution) Let v be any vector pointing down the incline, and let F g be the force on the wagon due to gravity. The magnitude of this force is F g = 40g newtons, where g = 9.8 represents the acceleration due to gravity. We can decompose this force into two vectors, one of which is parallel to the incline and one of which is perpendicular, as seen in the following figure: Here F v is the projection of F g onto v, and F v = F g F v is the normal force perpendicular to the incline. The force that results from pushing the wagon up the incline needs only to counteract the force F v. Because the angle between F v and F g is 90 10 = 80, F v has has magnitude F v = proj v F g = F g cos(80 ) 40 9.8 0.1736 68.07N. So a force of 68.07 newtons must be applied to push the wagon up the incline (or at least to hold the wagon in place). An example unrelated to dot products Example 5. (Paraphrased from 13.2, Exercise 67 of the textbook) Consider the triangle ABC in the plane. A median of the triangle is a line segment connecting a vertex of the triangle to the midpoint of the opposite side. The three medians intersect in a point called the centroid of the triangle. Show that if the points A, B, and C are represented by the vectors v, w, and u, respectively, then the centroid is P = 1 3 (u + v + w). (Solution) See the textbook for the associated figure. Denote by A the midpoint of the segment BC, opposite A. Following the hint in the text, we want r A (t) to parametrize the line segment connecting A to A. Because A is the midpoint of BC, we may represent it as a vector by w + 1 2 (u w) = 1 (u + w). 2 (The expression on the left starts at the point B and moves half the distance to C, since the vector from B to C is u w. The expression on the right just averages the vectors representing B and C. Either of these is a good way to think of the midpoint.) The line segment we want to parametrize should start at A and move towards A. If we re writing points as vectors we can write this as r A (t) = A + t(a A), 5
since A A should be the vector pointing from A to A. But we have vectors representing the points A and A and substituting these in gives us ( ) 1 r A (t) = v + t (u + w) v = v + t (u + w 2v). 2 2 Now we want to show that the point P lies on the line segment between A and A. We can do this by noticing that 1 r A (2/3) = v + 1 2 2 3 (u + w 2v) = v + 1 3 u + 1 3 w 2 3 v = 1 (u + v + w) = P. 3 We don t particularly care right now that 2/3 is the t-value for which r A (t) = P ; what s really important is that P lies on the line traced out by r A (t) that is, it lies on the median connecting A to A. Some similar trickery shows that P also lies on the medians connecting B to B and C to C. Since P lies on all three medians, it must be the centroid. 1 Where did the value 2/3 come from?! We want to show that P lies on the line traced out by r A(t), which means we need to find a t-value for which r A(t) = P. We can do this with some algebra: v + t 2 (u + w 2v) = 1 (v + u + w) 3 6v + 3t(u + w 2v) = 2(v + u + w) 3t(u + w 2v) = 4v + 2u + 2w = 2(u + w 2v). From this last line we see that letting t = 2/3 will do the trick. 6