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M243. Fall 2011. Homework 4. Solutions

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M243. Fall 2011. Homework 4. Solutions M243. Fall 2011. Homework 4. Solutions. H4.1 Given a cube ABCDA1B1C1D1, with th all sides of length 1, and with sides AA1, BB1, CC1 and DD1 being parallel (can think of them as \vertical"). (i) Let M denote the center of the square ABCD and let N be a point of the segment BB such that BN = 3 . Find the distance between lines MC and AN. 1 NB1 2 1 (ii) Let M denote the center of the square ABCD and let N be a point of the segment BB such that BN = 3 . Find the distance between M and the plane passing through 1 NB1 2 points A1, N, and C. Solution (i) We introduce the coordinate system such that A(0; 0; 0), D(1; 0; 0), B(0; 1; 0), and A1(0; 0; 1). Then C1(1; 1; 1), M(1=2; 1=2; 0) and N(0; 1; 3=5). To find the distance d between two lines one can use the following idea. The distance is the length of the segment which joins a point on one line to a point to another line and is perpendicular to each of them. Let ~n be any nonzero vector perpendicular to both lines and let PQ be an arbitrary segment with endpoints on the lines. Then −−! d = j comp~nPQ j −−! −−−! −−! −−! Let ~n = AN × MC1 = h0; 1; 3=5i × h1=2; 1=2; 1i = h7=10; 3=10; −1=2i, and PQ = AC1 = h1; 1; 1i. Then −−! h7=10; 3=10; −1=2ih_1; 1; 1i 1=2 p d = j comp PQ j = j j = jp j = 5 83=83 ≈ 0:5488 ~n jh7=10; 3=10; −1=2ij 83=10 (ii) To find the distance d between a point, P , and a plane, we use the fact that d is equal to the length of the line segment connecting a point on the plane to P and perpendicular to the plane. Thus if ~n is any nonzero vector normal to the plane determined by the points A1, N and C, and A1M is the line segment extending from A1 to M, then we have −−−! d = j comp~nA1M j: −−−! Now A1M = h1=2; 1=2; −1i, and we can take ~n as −−! −−! ~n = A1N × A1C = h0; 1; −2=5i × h1; 1; −1i = h−3=5; −2=5; −1i Then 1 −−−! −−−! j~n · A M j j comp A M j = 1 = ~n 1 j ~n j p jh1=2; 1=2; −1i · h−3=5; −2=5; −1ij 1=2 5 38 = p = ≈ 0:40555: j h−3=5; −2=5; −1i j 38=5 76 H4.2 Find equations of the planes that are parallel to the plane x + 2y − 2z = 1 and two units away from it. Solution It is clear that there have to be two such planes in the answer. Since planes are parallel, we can assume that the planes we are looking for have the same normal vector as the given plane. Hence, equation of each of them can be written in the form α : x + 2y − 2z + d = 0, where d has to be determined. The distance between parallel planes can be found as the distance from a point of one of them to another. Note that P : (1; 0; 0) is in the given plane. By the known formula: j 1 + 2 · 0 − 2 · 0 + d j dist (P; α) = = 2: p12 + 22 + (−2)2 This is equivalent to j1 + dj = 2 , d = −7 or d = 7: 3 Hence α : x + 2y − 2z − 7 = 0 or α : x + 2y − 2z + 2 = 0. (Note that instead of P we could use any other point in the plane x + 2y − 2z = 1.) In the same way one can establish a more general useful fact: the distance D between two parallel planes with equations ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by the formula jd − d j D = p 1 2 a2 + b2 + c2 Using this formula we would obtain: jd − (−1)j 2 = p12 + 22 + (−2)2 Solving for d gives d = 5 or d = −7. H4.3 It is known from geometry that three medians in a triangle are concurrent, and that three altitudes in a triangle are concurrent. In one previous homework we discussed an analogous property of a tetrahedron, namely that four medians of a tetrahedron are concurrent, and that their common point divides them in proportion 3:1. 2 An altitude of a tetrahedron is a line which passes through its vertex and perpendicular to the plane containing three other vertices (i.e., the opposite face). Are all four altitudes of an arbitrary tetrahedron concurrent? If your answer is YES, prove it. If your answer is NO, present a counter example (i.e., a tetrahedron where this is not true), and justify it. Solution This claim is false, which we will demonstrate with a counter-example. Consider the 3 tetrahedron in R with vertices A = (0; 0; 0), B = (0; 0; 1), C = (1; 0; 0), and D = (1; 1; 0). Let E be a point on line AD such that segment CE is perpendicular to line AD. Note that E is different from A. Obviously, the line AB (z-axis) is perpendicular to the plane defined by points A; C; D (xy-plane). So it is the altitude of the tetrahedron at vertex B. We know that if a line is perpendicular to two nonparallel lines in the plane, then it is perpendicular to the plane. Therefore line CE, being perpendicular to lines AB and AD, is perpendicular to the plane defined by points A; B; and D. So line CE coincides with the altitude of the tetrahedron at C. Obviously, lines AB and CE are skew lines, hence they do not intersect. Therefore the altitude of the tetrahedron are not concurrent. Comment. Tetrahedra when altitudes are concurrent are rare. One can explain that they can be characterized by the property that in them all three pairs of opposite (skew) edges are perpendicular. Hence, if one picks an arbitrary tetrahedron, with very high probability its altitudes are not concurrent. H4.4 Let ABCD be a tetrahedron. Let ~nA; ~nB; ~nC ; ~nD be vectors of unit length orthogonal to the faces 4BCD; 4ACD; 4ABD; 4ABC, respectively, and directed outwards of the tetrahe- dron. Prove that area (4BCD) ~nA + area (4ACD) ~nB + area (4ABD) ~nC + area (4ABC) ~nD = ~0: Each addend above is the product of a scalar (the area of of a triangle) by the corresponding unit vector. This property illustrates an interesting fact of physics. Do you see which one? (Hint: review the properties of cross product of vectors. ). −−! −! −−! Solution 1. Let ~a, AB = ~b, AC = ~c, and AD = d~. We know that the area of a triangle can be expressed as half of the magnitude of the cross product of the vectors defined by its sides. Moreover, the vector normal to the face of the triangle can be expressed in terms of this cross product as well. Throughout this process, it is very important that the correct order of the cross products be used. To determine this, it is perhaps best to draw a picture with labeled vertices and be consistent using the right hand rule to figure out which way the cross product should point. This way we obtain −−! −−! BD × DC area (4BCD) ~n = : A 2 3 Similarly, it follows that −! −−! AC × CD area (4ACD) ~n = ; B 2 −−! −−! AD × DB area (4ABD) ~n = ; and C 2 −−! −−! AB × BC area (4ABC) ~n = : D 2 Now, rewriting these in terms of the vectors ~b, ~c, and d~, and adding them together, we obtain area (4BCD) ~nA + area (4ACD) ~nB + area (4ABD) ~nC + area (4ABC) ~nD = −−! −−! −! −−! −−! −−! −−! −−! BD × DC AC × CD AD × DB AB × BC + + + = 2 2 2 2 (d~ −~b) × (~c − d~) ~c × (d~ − ~c) d~ × (~b − d~) ~b × (~c −~b) + + + = 2 2 2 2 1 (d~ −~b) × (~c − d~) + ~c × (d~ − ~c) + d~ × (~b − d~) +~b × (~c −~b): 2 Now we just use distributive property of the cross products inside the brackets, and the facts that ~x × ~y = −~y × ~x and ~x × ~x = ~0 for any ~x;~y. We obtain (d~ −~b) × (~c − d~) + ~c × (d~ − ~c) + d~ × (~b − d~) +~b × (~c −~b) = d~c~ −~b~c − ~0 +~bd~ + ~cd~ − ~0 + d~~b − ~0 +~b~c = (d~c~ + ~cd~) + (−~b~c +~b~c) + (~bd~ + d~~b) = ~0 + ~0 + ~0 = ~0: 1 ~ ~ Thus, the original sum is equal to 2 0 = 0, as we were trying to show. Solution 2. This can also be done with Maple by using the components of vectors. Physics Interpretation: The beautiful thing about this problem is that under the appro- priate physical interpretation, the answer is immediately clear without any calculation. Suppose our tetrahedron is a rigid container with some gas in it. Then since the pressure is constant throughout, we know the gas will exert a force normal to each face of the solid exactly proportional to the area of that face. However, we know that if our solid is initially placed at rest, then by conservation of momentum, it will not move (and therefore, it will not accelerate). Thus, all of the forces must balance each other, and their sum as vectors must be zero, completing the proof. 4 Furthermore, with this physics interpretation, it is immediately clear that this can be generalized to any polyhedron whatsoever, and that it can even be extended to polygons 2 3 in R as the limiting case of solids in R . 5.
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