Answers and Solutions to Section 9.3 Homework S. F. Ellermeyer

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1 Answers and Solutions to Section 9. Homework S. F. Ellermeyer. (a) (a b) c is meaningless because a b is a scalar and c is a vector. We can t take the dot roduct of a scalar and a vector. (b) (a b) c is meaningful. It is a vector that is a scalar multile of c. (c) jaj (b c) is meaningful. It is a scalar. (d) a (b + c) is meaningful. It is a scalar. (e) (a b) + c is meaningless because a b is a scalar and c is a vector. We can t add a scalar and a vector. (f) jaj(b + c) is meaningless because jaj is a scalar and b+c is a vector.. a b jaj jbj cos () (6) (5) 5 5. a b (4) (6) + () ( ) + 4 ( 8) 9 7. a b () (5) + ( ) (0) + () (9) 9. If we assume that the triangle that is ictured is equilateral, then we have u v juj jvj cos (60 ) () () u w juj jwj cos (0 ) () (). 0. u v juj jvj cos (45 ) () u w juj jwj cos (90 ) () () (0) 0.. Here are two ways to show that, for examle, i k 0. First note that the angle between i and k is 90. Thus ik jij jkj cos (90 ) () () (0) 0: The other way of showing it is i i + 0j+0k and k 0i + 0j+k so i k () (0) + (0) (0) + (0) () 0.. For a h 8; 6i and b 7; cos () a b jaj jbj ( 8) 7 + (6) () q q 7 ( 8) Thus, 0:0795 arccos ( 0:0795) 94:54.

2 5. For a h0; ; i and b h; ; i cos () a b jaj jbj (0) () + () () + () ( ) q + + ( ) Thus, 8 8 : arccos 8 00: For a h 5; ; 7i and b h6; 8; i, we have a b ( 5) (6) + () ( 8) + (7) () 40 which means that angle between a and b is obtuse (and hence a and b are not erendicular). Also and jaj jbj q ( 5) q 6 + ( 8) + 6 from which it can be seen that a b 6 jaj jbj. Hence 6 80 : We conclude that a and b are neither arallel nor erendicular.. Let a xi+yj+zk be the vector we are seeking. Since we want this vector to be erendicular to i + j, we must have a (i + j) 0. This gives us (xi + yj + zk) (i + j) x + y 0:

3 Likewise, since we want a vector to be erendicular to i + k, we must have a (i + k) 0. This gives us (xi + yj + zk) (i + k) x + z 0: A solution of the system of equations x + y 0 x + z 0 is x, y, z. Thus, we can take a i + j + k. To check that a is erendicular to both i + j and i + k, note that and 7. We de ne and note that ( i + j + k) (i + j) 0 ( i + j + k) (i + k) 0. orth a b b roj a b b b a jaj a aorth a b a b b a jaj a a b b a jaj a a a b b a jaj jaj 0. This shows that orth a b is orthogonal to a. Also note that roj a b+orth a b b.

4 9. Let a h; 0; i. We want to nd a vector b such that com a b. Suose that b hx; y; zi. Then we want to have b a jaj. x z 0 x z 0. Thus, for examle, we could take b 0; 0; The dislacement vector is D h4 ; 9 ; 5 0i h; 6; 5i and the force vector is F h0; 8; 6i. The work done by the force is W D F () (0) + (6) (8) + (5) ( 6) 8 Joules.. The work done is W D F jdj jfj cos (0) (0) (5) cos (0 ) 5 foot ounds. 5. Let P 0 (x 0 ; y 0 ) be the oint on the line ax+by +c 0 that is closest to the oint P (x ; y ). Let P (x ; y ) be any oint on the line ax + by + c 0 such that P 6 P 0. (See the gure below). 4

5 First, we note that the vector ha; bi is orthogonal to the line ax+by+c 0. To see why this is true, observe that ha; bi P 0 P ha; bi hx x 0 ; y y 0 i a (x x 0 ) + b (y y 0 ) ax + by (ax 0 + by 0 ) c ( c) 0. The distance from the oint P (x ; y ) to the line ax + by + c 0 is the distance from the oint P (x ; y ) to the oint P 0 (x 0 ; y 0 ). This, in turn, is equal to the absolute value of the scalar rojection of the vector P P 5

6 onto the vector ha; bi. Thus, the distance from the oint to the line is P P ha; bi roj ha;bi P P jha; bij hx x ; y y i ha; bi a + b a (x x ) + b (y y ) a + b ax + by (ax + by ) a + b j c ax by j a + b jax + by + cj a + b. 7. Picture the cube to have one of its vertices at the oint (0; 0; 0) and the three edges that come together at the vertex to lie in the three coordinate axes. If each edge of the cube has length d, then the diagonal of the cube that oints from (0; 0; 0) to (d; d; d) is a reresentative of the vector r di + dj + dk. If is the angle between the vector r and, say, the vector i (which is arallel to one of the edges of the cube), then cos () r i jrj jij d d. Therefore arccos 54: CH ; 0 ; 0 ; ; and CH ; ; ; ;. If is the angle between CH and CH, then CH CH cos () CH CH 4 4. Thus arccos 09:5. 4. ja bj jjaj jbj cos ()j jaj jbj jcos ()j jaj jbj () jaj jbj. 6

7 44. The geometric interretation of the triangle inequality, ja + bj jaj + jbj, is that the length of any side of a triangle must be less than the sum of the lengths of the other two sides. (See the icture.) We rove the triangle inequality as follows: ja + bj (a + b) (a + b) a (a + b) + b (a + b) a a + a b + b a + b b jaj + a b+ jbj jaj + ja bj + jbj jaj + jaj jbj + jbj (jaj + jbj). (Note that the Cauchy Schwarz inequality, ja bj jaj jbj, was used.) Having roved that ja + bj (jaj + jbj), we obtain ja + bj jaj + jbj. 45. The geometric interretation of the Parallelogram Law, ja + bj +ja bj jaj + jbj, is that the sum of the squares of the lengths of the diagonals of a arallelogram is equal to the sum of the squares of the sides of the arallelogram. (See the Figure.) 7

8 We rove the Parallelogram Law as follows: ja + bj + ja bj (a + b) (a + b) + (a b) (a b) jaj + a b+ jbj + jaj a b+ jbj jaj + jbj. 8