M243. Fall Homework 2. Solutions.

Transcription

1 M43. Fall 011. Homework. s. H.1 Given a cube ABCDA 1 B 1 C 1 D 1, with sides AA 1, BB 1, CC 1 and DD 1 being parallel (can think of them as vertical ). (i) Find the angle between diagonal AC 1 of a cube and diagonal AB 1 of its face (ii) Repeat part (i) replacing AB 1 by A 1 B (iii) Let M denote the center of the square ABCD and let N be a point of the segment BB 1 such that BN NB 1 3. Find the angle between lines MC 1 and AN. We introduce the coordinate system such that A(0, 0, 0), D(1, 0, 0), B(0, 1, 0), and A 1 (0, 0, 1). In each part we denote the measure of the unknown angle between the corresponding vectors by γ. As we know, if γ [0, π/], then the angle between the corresponding lines is γ. If γ (π/, π], it is π γ. (i) Since C 1 (1, 1, 1) and B 1 (0, 1, 1), then AC 1 1, 1, 1, and AB 1 0, 1, 1. Hence cos γ AC 1 AB 1 / AC 1 AB 1 /( 3 ) , and γ radians. (ii) Since A 1 (0, 0, 1), then A 1 B 0, 1, 1. Therefore cos γ AC 1 A 1 B/ AC 1 A 1 B 0, and γ π/ radians. (iii) Coordinates of points M and N are easy to find from a drawing, but general formuli from H1.1 (ii) can be used if it creates difficulties. One gets M(1/, 1/, 0) and N(0, 1, 3/5). Therefore MC 1 1/, 1/, 1, and AN 0, 1, 3/5. Hence cos γ MC 1 AN/ MC 1 AN 1.1/( ) , and γ radians. H. Let A 1, A, A 3, A 4 be four points in R 3. Suppose lines A 1 A and A 3 A 4 are perpendicular, and lines A 1 A 3 and A A 4 are perpendicular. Prove that then lines A 1 A 4 and A A 3 are also perpendicular. To simplify the notations, denote vectors A 1 A i by e i, i, 3, 4. Then the given perpendicularity conditions can be rewritten as A 1 A A 3 A 4 A 1 A A 3 A 4 0 e ( e 4 e 3 ) 0 e e 4 e e 3 (1) Similarly (or just interchanging indices and 3), we have A 1 A 3 A A 4 e 3 e 4 e 3 e () We want to show that A 1 A 4 A A 3 e 4 e 3 e 4 e (3) Obviously (1) and () imply (3) and we are done. 1

2 H.3 (i) If c a b + b a, where a, b, c are all nonzero vectors, show that c bisects the angle between a and b. (ii) Let OA, OB, OC be three rays in space, and let rays OD, OE, OF bisect the angles AOB, BOC, COA, respectively. Prove that the three angles formed by the rays OD, OE, OF are either all acute, or all right, or all obtuse. (i) Since a b a b b a b a, then the parallelogram with sides determined by vectors a b and b a is a rhombus and c corresponds to its diagonal. But a diagonal of a rhombus bisects its angle: the obtained two triangles are congruent by SSS. Clearly the same argument gives a more general statement: the sum of two vectors of equal length bisects the angle between them. The proof is finished. Suppose one wants to prove that the sum z x + y of two vectors x and y of equal lengths forms equal angles with each of them without using geometry. It also can be easily done. Let α denote the measures of the angle between x and z, and β denote the measures of the angle between y and z. Since x y, then cos α x z x x + x y x z x z x + x y x z y + x y y z y x + y y y z y z y z cos β By definition of the angle between two vectors, both α and β are in [0, π]. Therefore the equality of their cosines, implies the equality of the angles, and the proof is finished. (ii) The statement of the problem can be rephrased as the following: show that the cosines of the angles determined by the three rays are either all positive, or all zero, or all negative. It is also clear that the sign of the cosine of the angle between two vectors is determined by the sign of their dot product: x y cos γ x y, and the denominator is always positive for nonzero vectors. This suggest the following solution. Consider three unit vectors: e A 1 1 OB OB, and ec 1 OC. OC OA OA, e B Since e A, e B, e C are all of equal length ( 1 ), vectors e A + e B, e B + e C and e C + e A are parallel to rays OD, OE, OF and bisect the angles AOB, BOC, COA, respectively, as was explained in part (i). Therefore the sign of cos( DOE) is determined by the sign of ( e A + e B ) ( e B + e C ) e A e B + ( e B ) + e B e C + e C e A 1 + e A e B + e B e C + e C e A Since the final expression of the dot product above is symmetric with respect to all permutations of letters A, B, C, the signs of cosines of two other angles will be determined by the sign of exactly the same expression! (In case you have doubts, just repeat the computation for two other angles). This ends the proof.

3 H.4 Suppose points A 1, A,...A n lie on a unit circle centered at point O and divide it into n congruent arcs. (i) Find A 1 A i (ii) Find the sum of squares of lengths of all segments A i A j, 1 i < j n. (Hint: Use Problem H1.4) (i) Let O be the center of the circle and e i OA i. Then A 1 A i e i e 1 and by the statement of Problem H1.4, we know that n e i 0. Therefore A 1 A i ( e i e 1 ) ( e i e 1 ) n e 1 Therefore the answer is n. ( e i e i e 1 + e 1 ) e i n e 1 0 n 0 n (ii) It is clear from the symmetry of the regular n-gon, that replacing the vertex A 1 in the the sum n j1 A 1 A j by any other vertex A i, leads to the same answer n. Therefore 1 j n A ia j n. Hence we have A i A j 1 A i A j 1 A i A j 1 i<j n 1 1 i j n 1 i n 1 i n (n) 1 n(n) n 1 j n The constant 1/ in front of the second sum above is due to the fact that every addend from the first sum appears in the second sum exactly two times. H.5 Problem # 38 from Section 13.4 We know that u v u v sin γ (3)(5) sin γ 15 sin γ, where γ is the measure of the angle between u and v. Since by definition γ [0, π], the maximum value of the u v is 15 and the minimum is 0. Since u rotates in the xy-plane and v 5 j, the direction of u v is always parallel to z-axis (unless it is 0). Whether it is positive or negative direction of the z-axis, depends on the quadrant in the xy-plane where the endpoint of u lies. It is positive if it is in the 4-th or the 1-st quadrant, and it is negative in the -nd and the 3-rd quadrant. 3

4 H.6 Problem # 4 from Section (If you wish, you can use Maple to solve this problem). This is a good exercise for Maple. Our goal is to prove the property of the triple cross product of vectors a x (b x c) (a dot c) b - (a dot b) c (*) It is clear that this can be done by using components of vectors a,b,c. We start with loading the package linalg which contains commands related to vectors and matrices. To supress the display of the long list of commands containing in the package we end the line with : > with(linalg): Next we write the vector by using components > a: vector( [a1,a,a3] ); b : vector( [b1,b,b3] ); c: vector( [c1,c,c3] ); a : [a1, a, a3] b : [b1, b, b3] c : [c1, c, c3] The next line computes the left hand side of (*), ie a x (b x c): > LHS : simplify( crossprod(a, crossprod(b,c))); LHS : [a b1 c - a b c1 - a3 b3 c1 + a3 b1 c3, a3 b c3 - a3 b3 c - a1 b1 c + a1 b c1, a1 b3 c1 - a1 b1 c3 - a b c3 + a b3 c] The following three lines compute the right hand side of (*): > Y : simplify( scalarmul(b,dotprod(a,c, orthogonal )) ); Y : [(a1 c1 + a c + a3 c3) b1, (a1 c1 + a c + a3 c3) b, (a1 c1 + a c + a3 c3) b3] > Z : simplify (scalarmul(c,(-1)*dotprod(a,b, orthogonal )) ); Z : [-(a1 b1 + a b + a3 b3) c1, -(a1 b1 + a b + a3 b3) c, -(a1 b1 + a b + a3 b3) c3] > RHS : simplify( matadd(y,z)); RHS : [a b1 c - a b c1 - a3 b3 c1 + a3 b1 c3, a3 b c3 - a3 b3 c - a1 b1 c + a1 b c1, a1 b3 c1 - a1 b1 c3 - a b c3 + a b3 c] Now we compare the expressions on both sides of (*) by subtracting them: 4

5 > simplify( matadd(lhs, scalarmul(rhs,-1))); [0, 0, 0] Since the result is zero vector, LHS RHS, and the statement is proven. 5