LB 220 Homework 2 (due Tuesday, 01/22/13)

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Ami Kennedy
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1 LB 220 Homework 2 (due Tuesday, 01/22/13) Directions. Please solve the problems below. Your solutions must begin with a clear statement (or re-statement in your own words) of the problem. You solutions should be clear, legible, and demonstrate at minimum partial progress towards a complete solution to the problem. Please refer to the syllabus for the policy on grading (communication, completeness, and correctness) and late homework (homework is due at the start of class, late homework is assessed a 20% penalty if submitted within the next 48 hours) Collaboration. I encourage you to discuss the homework problems with your classmates. However, each student must write and submit his or her own homework solutions. Calculators. You can use calculators to determine a numerical approximation to an answer to an application question, but you should use exact values until the very last step in the problem. Calculators are not, however, permitted on any quizzes or exams. 1. A boat sails south with the help of a wind blowing in the direction S36 E with a magnitude of 400 lb. Find the work done by the wind as the boat moves 120 ft. Solution: The equation for work W is W = F P Q, where F is the force applied to an object which moves from P to Q. The boat is moving moving from an initial point, which may be chosen to be the origin O(0, 0) to a terminal point Q(0, 120) which is 120 ft due south. The magnitude of the force applied is 400 lb (the pound here refers to units of force the pound-force lb f, to be precise; for more on this, please read the solution to exercise #5 on the first homework.) Using the physicists definition of the scalar product, we have that Work = F P Q cos 36 = (400)(120) cos 36 38, 833 ft lb. The answer in the metric system, using 400 lb 4.45 N and 1 ft m, is approximately equal to 5, 2671 N m or, equivalently, 5, 2671 J. 2. Which of the following expressions is meaningful? If it is, state whether the result is a vector or a scalar; if it is not, explain why.

2 (a) a (b c) (b) a (b c) (c) a (b c) (d) a (b c) (e) (a b) (c d) (f) (a b) (c d) Solution: (a) is meaningful, the result is a scalar (b) is meaningless since the computation involved the vector product of a vector and a scalar (c) is meaningful, the result is a scalar, though be careful with the vector product it is not associative (d) is meaningless since the computation involves the scalar product of two scalars; however, if all three vectors are 1-dimensional vectors (i.e. vectors in R 1, then the expression is meaningful and simply is the product of these three numbers. (e) is meaningless since the computation involves the vector product of two scalars (f) is meaningful, the result is a scalar 3. Determine the two unit vectors which are orthogonal to the plane through the points P (1, 0, 1), Q( 2, 1, 3), and R(4, 2, 5). Then compute the area of triangle P QR. Solution: The vector P Q P R is orthogonal to the plane containing P QR: P Q i j k P R = = 0, 18, The magnitude of this vector is the area of the parallelogram spanned by the two vectors; half this value is the area of the triangle. So, the area of the triange is equal to = = The two unit vectors orthogonal to the plane are , 18, 9 = 0, 2 5, 1 5 and its negative.

3 4. Prove the following statement: if P is a point not on a line L that passes through the points Q and R, then the distance from P to L is equal to a b / a, where a is represented by QR and b by QP. Solution: The distance d from P to L is equal to the length of a line segment joining P to L so that the segment is orthogonal to L. Suppose this line segment meets L in a point called N. Let θ be the measure of the angle P QN. Then d = QP sin θ. (There are two cases: either the angle is acute or the angle is obtuse, but since sin (π θ) = sin θ both cases yield the same equation.) Next cleverly multiply by 1: d = QR QP sin θ = QR QP QR QR, the last inequality following from the physicists definition of the vector product. There is one small tecnical point: what do we do if N = P? In this case, we simply use R in place of P and everything works out in an analogous way. 5. Find a vector equation and parametric equations for the line of intersection of the planes x + 2y + 3z = 1 and x y + z = 4. Solution: There are several approaches to this problem. Here is one approach. If P (0, y, z) is a point common to both planes, then 2y + 3z = 1 and y + z = 4. Multiplying the second equation by 2, adding the result to the first equation, and solving for z yields z = 9/5. Substituting back in to either equation yields y = 11/5. Therefore, P (0, 11/5, 9/5) is a point in the intersection of the two planes. We could do this again starting with some point Q(1, y, z) so that 1 + 2y + 3z = 1 and 1 y + z = 4. Solving this system of two equations reveals that Q(1, 9/5, 6/5). Therefore, a vector equation of the line of intersection is r = OP + tp Q: r = 0, 11/5, 9/5 + t 1, 2/5, 3/5, which is equivalent to x = t, y = 11/5 + 2t/5, z = 9/5 3t/5. Another approach is to stop after finding the point P and then use the vector product of normal vectors to the planes to determine a

4 vector parallel to the line of intersection: i j k v = = 5, 2, So, the line has equation r = OP + tv: r = 0, 11/5, 9/5 + t 5, 2, 3, which is equivalent to x = 5t, y = 11/5 + 2t, z = 9/5 3t. A third approach is to set x = t and to solve the pair of equations t + 2y + 3z = 1 and t y + z = 4 for y and z in terms of t. As before, mulitply the second equation by 2 and add it to the first. This yields 3t + 5z = 9. Therefore, z = 3t/5 + 9/5. Substiting back into either equation yields y = 2t/5 11/5. It may seem odd that there can be two such very different looking equations for the same line. Parametrizations are not unique. We can check that these are indeed the same line by checking whether P Q and v are parallel. Indeed they are: 5 P Q = v. 6. Determine an equation of the plane which contains the line x = 1 + t, y = 2 + t, z = t and which is parallel to the plane x + y + 2z = 5. Solution: The approach is to use the normal vector 1, 1, 2 of the plane and any point on the line. For instance, if t = 0, then P (1, 2, 0) lies on the line. Therefore, the following is an equation of a plane parallel to the given plane and containing P : (x 1) + (y + 2) + 2z = 0. Finally, we need to verify that the line is indeed contained in this plane. You can either substitute in the equations: (1 + t 1) + ( 2 + t + 2) + 2( t) = 0 and check it really equals zero. Or you might verify that the vector 1, 1, 1 which is parallel to the line is orthogonal to the normal vector 1, 1, 2 : 1, 1, 1 1, 1, 2 = 0.

5 7. Determine the cosine of the angle between the planes 5x + 2y + 3z = 0 and x y + z = 2. Solution: The cosine of the angle between the planes is the same as the cosine of the angle between a pair of normal vectors, each corresponding to one of the planes. 5, 2, 3 1, 1, 1 = = 6 On the other hand, using the physicists definition of the dot product, the above is equal to cos θ. Therefore cos θ = = Let L 1 be the line through the origin and the point (1, 1, 1). Let L 2 be the line through the points (2, 3, 4) and ( 1, 1, 0). Determine the distance between L 1 and L 2. Solution: The strategy is as follows. First, observe that the shortest line segment joining a point of L 1 to a point of L 2 must meet both lines at right angles. So, we first find a vector orthogonal to both lines by using the vector product: i j k n = = 0, 1, Next, write an equation for the plane containing L 1 and normal to n. For instance, using the point O(0, 0, 0) of L 1, the following is such an equation: y + z = 0. Finally, choose any point of L 2, for instance P (2, 3, 4) and compute the distance from P to the plane above. This distance is the same as the distance between the two lines. (This may not be clear, but if you try visualizing this with a notebook as a prop for a plane and a pair of pens as props for the lines, you will soon be convinced this is correct.) Using the formula from from the book, this distance d satisfies d = OP n = 1. n 2

6 Remark: Here is a nice formula that I learned from a student: if a is a vector parallel to a line L and b a vector parallel to a line M and if P is a point on L and Q is a point on M, then if L and M are skew lines then the distance between L and M is given by P Q (a b). a b The proof follows the same outline as in the problem above. You need to be careful with such a formula, however: it does not give the correct answer if the line are not skew. For example, if the lines are parallel, then the above quantity is undefined (since the cross product in the denominator is zero). Similarly, if the lines intersect then their distance is zero, but the formula above is non-zero. 9. Sketch the traces of the surface x 2 + y 2 z 2 = 1 and explain why the surface looks like a hyperboloid of one sheet. If the equation is changed to x 2 y 2 + z 2 = 1, how is the surface changed? What if it is changed to x 2 + y 2 + 2y z 2 = 0? Solution: The trace of the surface in the plane z = c has equation x 2 + y 2 = 1 + c 2, which is a circle of radius 1 + c 2. The curves x 2 + y 2 = 1 + c 2 for c = 0, 1, 2, 3, 4.

7 The trace of the surface in yz-plane has equation y 2 z 2 = 1, which is a hyperbola. The curve y 2 z 2 = 1. The trace of the surface in the xz-plane has equation x 2 z 2 = 1, which is a hyperbola. From this we conclude that the surface is a hyperboloid. It is a hyperboloid of one sheet because for every plane z = c, the trace is a circle. If it were a two-sheeted hyperboloid, then one of these planes would not intersect the surface at all. If the equation is changed to x 2 y 2 + z 2 = 1, then the roles of y and z have been interchanged. This is realized by a reflection in the plane y = z in R 3. So, the resulting surface has the same shape (it is still a one-sheeted hyperboloid), but the y-axis is now the radial axis. If the equation is changed to x 2 + y 2 + 2y z 2 = 0, then the hyperboloid is translated from being centered at the origin to being centered at the point (0, 1, 0). This can be seen by completing the square: x 2 + (y + 1) 2 z 2 = Sketch the traces of the surface x 2 y 2 + z 2 = 1 and explain why the surface looks like a hyperboloid of two sheets. If the equation is changed to x 2 y 2 z 2 = 1, how is the surface changed? What if it is changed to x 2 + 2x y 2 z 2 = 0?

8 Solution: The trace of the surface in the plane z = c has equation x 2 + y 2 = c 2 1, which is a circle of radius c 2 1 if c 2 > 1. If c 2 = 1, then the trace is a single point. if c 2 < 1, then the trace is empty. For this reason this will be a hyperboloid of two-sheets. As in the problem above, the traces in the xz- and yz-planes are hyperbolas. If the equation is changed to x 2 y 2 z 2 = 1, then the roles of x and z have been interchanged. This is realized by a reflection in the plane x = z in R 3. So, the resulting surface has the same shape (it is still a two-sheeted hyperboloid), but the x-axis is now the radial axis. If the equation is changed to x 2 + 2x y 2 z 2 = 0, then the hyperboloid in the previous paragraph (having the x-axis as its radial axis) is translated from being centered at the origin to being centered at the point ( 1, 0, 0). This can be seen by completing the square: (x + 1) 2 y 2 z 2 = 1. So, for instance, the two points at the base of the sheets are ( 1, 0, 0) and (1, 0, 0).

LB 220 Homework 1 (due Monday, 01/14/13)

LB 220 Homework 1 (due Monday, 01/14/13) Directions. Please solve the problems below. Your solutions must begin with a clear statement (or re-statement in your own words) of the problem. You solutions