Brief Review of Exam Topics

Transcription

1 Math 32A Discussion Session Week 3 Notes October 17 and 19, 2017 We ll use this week s discussion session to prepare for the first midterm. We ll start with a quick rundown of the relevant topics, and then proceed to work examples and answer questions as requested. It s important to note that I m not writing the exam, so nothing presented here or in section should be taken as an indication of what you ll see on exam day. Brief Review of Exam Topics Vectors We first thought of vectors as a basepoint-terminal point pair, and denoted the vector from P to Q by P Q. We say that the vectors P 0 Q 0 and P 1 Q 1 are equivalent if they translate to one another. More simply, they have the same components. The algebraic operations of addition and scalar multiplication are defined on vectors. Graphically we have: In terms of components, these operations are all done component-wise. The magnitude of a vector is defined to be the distance from its basepoint to its terminal point. In components we have a, b, c = a 2 + b 2 + c 2. A linear combination of the vectors v and w is a vector av + bw, where a and b are scalars. If we restrict the scalars to the range 0 a, b 1, the set of linear combinations is called the parallelogram spanned by v and w. Parametrized Lines We describe the line that passes through the point (x 0, y 0, z 0 ) in the direction of v = a, b, c by r(t) = x 0, y 0, z 0 + t a, b, c, 1

2 called a vector parametrization of our line, and also by x(t) = x 0 + at, y(t) = y 0 + bt, z(t) = z 0 + ct, called a set of parametric equations for our line. There are various parametrizations and sets of parametric equations for the same line. Given a parametrization, we often call the point associated to it the initial point and its vector the velocity vector. The speed of the parametrization is the magnitude of the velocity vector. The Dot Product Whenever two vectors have a common number of components, their dot product is defined and is a scalar. Specifically, if u = u 1,..., u n and v = v 1,..., v n, then u v = u 1 v u n v n. The dot product also has a very important geometric interpretation. If θ is the angle between u and v (we always assume that θ [0, π]), then Some important observations: u u = u u cos 0 = u 2 ; u v = 0 if and only if u v; u v = u v cos θ. u v > 0 if and only if θ is acute, and u v < 0 if and only if θ is obtuse. We can rewrite the above equation as a formula which allows us to compute θ: ( ) u v θ = arccos. u v We can also use the dot product to compute the projection of a vector u onto a line L that passes through the basepoint of u. If v is any nonzero vector on L, then the projection of u onto L is given by proj L u = u v v v v. We sometimes write proj L u as proj v u. The Cross Product The cross product is defined only for pairs of three-dimensional vectors. Also unlike the dot product, its output is another three-dimensional vector.the cross product of vectors v = v 1, v 2, v 3 and w = w 1, w 2, w 3 is the vector v w = v 1 v 2 v 3 w 1 w 2 w 3 = v 2 v 3 w 2 w 3 i v 1 v 3 w 1 w 3 j + v 1 v 2 w 1 w 2 k. 2

3 Like the dot product, we start with an algebraic definition, but we really prefer the geometric description. The cross product v w is the unique vector for which v w is orthogonal to both v and w; v w has length v w sin θ; {v, w, v w} forms a right-handed system. Some important observations: w v = v w; v w = 0 if and only if w = λv for some λ R or v = 0; the cross product distributes across sums. The cross product can be used to compute areas. If P is the parallelogram spanned by the vectors v and w, then Area(P) = v w. The cross product can be used alongside the dot product to compute volumes. If P is the parallelepiped spanned by the vectors u, v, and w, then Examples Vol(P) = u (v w). Example 1. ( 13.1, Exercise 69 of the textbook) Use vectors to prove that the diagonals AC and BD of a parallelogram bisect each other. (See figure.) (Solution) Let v = AB and let w = AD. Then C is at the terminal point of v + w, so 1(v + w) is the midpoint of AC. The vector DB is given by v w, so the midpoint of BD is w + 1 (v w), 2 obtained by starting at D and moving half the distance towards B. But w (v w) = 1 2 (v + w), so the midpoint of BD is the same as the midpoint of AC, meaning that the diagonals bisect each other. 2 Example 2. ( 13.2, Exercise 64) Find a vector parametrization for the line x 5 9 = y = z 10. 3

4 (Solution) We ll find two parametrizations: one that works, and one that s much nicer. The first approach is to just let x = t. This leads us to the equations y = t 5 9 y = 7 9 t and z 10 = t 5 z = t , which gives us a perfectly acceptable parametrization of the line: r(t) = 0, 62 9, t 1, 7 9, 1 9 We can find a more pleasant solution, though. Instead of letting x = t, let s choose x so that the expression (x 5)/9 simplifies to t. Setting (x 5)/9 = t yields x = 9t + 5. We then have (y + 3)/7 = t and z 10 = t, giving us This leads to the much nicer parametrization y = 7t 3 z = t r(t) = 5, 3, 10 + t 9, 7, 1. Assuming the common value to be t is generally a good strategy for turning symmetric parametric equations into a nice parametrization for a line. Example 3. ( 13.3, Exercise 92) Use the Cauchy-Schwarz inequality v w v w to prove the triangle inequality: v + w v + w. (Solution) We can write the square of v + w as. v + w 2 = (v + w) (v + w) = (v + w) (v + w). According to the triangle inequality for numbers we have (v + w) (v + w) = v v + 2v w + ww v v w + w 2. Applying the Cauchy-Schwarz inequality to the middle term on the right hand side, we see that v + w 2 = (v + w) (v + w) v v w + w 2 = ( v + w ) 2. So v + w 2 ( v + w ) 2. Since both v + w and v + w are non-negative, we may take the square root on both sides of this inequality to conclude that v + w v + w. Example 4. The vectors l, 0, 0, 0, w, 0, and 0, 0, h span a rectangular prism with length l, width w, an height h. Use the triple scalar product to verify that this rectangular prism has volume lwh. 4

5 (Solution) Write the vectors as l = l, 0, 0, w = 0, w, 0, h = 0, 0, h. Then the volume of the prism spanned by these vectors is V = l (w h). We first compute w h: Then as we suspected. w h = 0 w h = wh, 0, 0. V = l, 0, 0 wh, 0, 0 = lwh, Example 5. ( 13.4, Exercise 39) Compute the volume of the parallelepiped spanned by u = 1, 0, 0, v = 0, 2, 0, w = 1, 1, 2. (Solution) We have V = u (v w). Note that the order of our vectors in the triple scalar product doesn t affect the result, but it can affect how much work we do. Let s be super lazy. Since we see that the vectors u and v have two zeros each, let s make sure they show up in the cross product we have to compute. That is, we ll compute the volume using V = w (u v). We have so u v = = 0, 0, 2, V = 1, 1, 2 0, 0, 2 = 4. Example 6. ( 13.4, Exercise 45) Use cross products to find the area of the triangle in the xy-plane defined by (1, 2), (3, 4), and ( 2, 2). (Solution) Let s outline our approach: we ll fix one of the vertices of the triangle as a basepoint and consider the vectors v and w taking us to the other vertices. Treating these as vectors in three dimensions, we can use the cross product to compute the area of the parallelogram spanned by these vectors. The area of the triangle will be one half of this area. Treat (1, 2) as the basepoint, so that the vectors v = 2, 2 and w = 3, 0 span our triangle. Because the cross product is only defined in three dimensions, we ll actually write v = 2, 2, 0 and w = 3, 0, 0, since the z-component of these vectors is zero. Then Area(P) = v w, 5

6 so we compute v w = = 0, 0, 6, so v w = 6. We conclude that the triangle has area 3. Example 7. ( 13.4, Exercise 75) Show that if a, b are nonzero vectors such that a b, then there exists a vector X such that a X = b. (Solution) Let v = a b, and let w = a v. Then w is perpendicular to both a and v. But b is also perpendicular to both a (by assumption) and v (because v = a b). So w = λb for some scalar λ. Because v is perpendicular to a, the cross product w is nonzero, so λ 0. So we may let X = 1 λ v. Then a X = 1 λ a v = 1 λ w = b, just as we hoped. 6