Mathematics 220, Test 1 - Solutions F, 2010 Philippe B. Laval Name 1. Determine if each statement below is True or False. If it is true, explain why (cite theorem, rule, property). If it is false, explain why or give a counter example. Assume that u and v are D vectors. (7 points each) (a) u v v u This is False. The correct statement is u v v u (b) If u and v are unit vectors, so is u v. This is False. We know that u v u v sin θ where θ is the angle between the two vectors. If both vectors are unit vectors, we have u v sin θ. Thus, u v 1 only when the two vectors are orthogonal. (c) ( u v) u 0 This is True because u v is a vector perpendicular to u. (d) ( u + v) v u v This is True. Using the properties of the cross product, we get: ( u + v) v u v + v v u v + 0 u v 2. Calculate the given quantity if a 1, 1, 2, b, 2, 1. (7 points each) (a) The angle between a and b in degrees Let θ be the angle between the two vectors (recall that 0 θ π). Then we have: a b a b cos θ Therefore cos θ a b a b (1) () + (1) ( 2) + ( 2) (1) 1 2 + 1 2 + ( 2) 2 2 + ( 2) 2 + 1 2 1 6 14 So, ( ) 1 θ cos 1 84 θ 96
(b) The scalar projection of b onto a The scalar projection of b onto a is denoted comp a b and it is equal to comp a b a b a 1 6.408 (c) A unit vector orthogonal to both a and b A vector orthogonal to both a and b is a b. A unit vector orthogonal to both a and a b is b a. First, we compute b a b i j k 1 1 2 2 1 i 7 j 5 k So, a b i 7 j 5 k ( ) 2 + ( 7) 2 + ( 5) 2 8 Hence, a unit vector orthogonal to both a and b is 1 8, 7, 5. Find the value of x such that the vectors 2, x, 4 and 2x,, 7 are orthogonal. (7 points) These two vectors will be orthogonal if 2, x, 4 2x,, 7 0 that is 4x + x 28 0 7x 28 x 4 4. Find symmetric equations of the line through the point (0, 2, 1) which is parallel to x 1 + 2t the line with parametric equations y t (6 points) z 5 7t The symmetric equations of the line through (x 1, y 1, z 1 ) in the direction of a, b, c are x x 1 a y y 1 b 2 z z 1 c
In our case, the direction of the line is 2,, 7. Therefore, symmetric equations of the line through the point (0, 2, 1) in the direction of 2,, 7 are x 2 y 2 z + 1 7 5. Consider the two planes x + y z 1 and 2x y + 4z 5. (7 points each) (a) Show these planes are neither parallel nor perpendicular. Two planes are parallel if their normals are parallel. They are perpendicular if their normals are perpendicular. Let n 1 be the normal of the first plane and n 2 be the normal of the second plane. Then we have n 1 1, 1, 1 and n 2 2,, 4. Are the planes perpendicular? They will be if n 1 n 2 0. n1 n 2 1, 1, 1 2,, 4 2 4 5 We conclude the two planes are not perpendicular. Are they parallel? They will be if there exists a constant c such that n 1 c n 2. n1 c n 2 1, 1, 1 c 2,, 4 1, 1, 1 2c, c, 4c 2c 1 c 1 4c 1 c 1 2 c 1 c 1 4 Thus, the system has no solution. It follows the two vectors are not parallel and therefore the two planes are not parallel. Note: To find if the two vectors are parallel we could have also mused the cross product. Recall that two nonzero vectors are parallel if and only if their cross product is 0. Since 1, 1, 1 2,, 4 1, 6, 5 0, the vectors n 1 and n 2 are not parallel hence the planes are not parallel. (b) Find, correct to the nearest degree, the angle between these two planes. The angle between two planes is the acute angle between their normals. Let θ be the angle between the two planes. Then we have θ cos 1 n 1 n 2 n 1 n 2 cos 1 5 29 58
(c) Since the two planes are not parallel, they intersect. Let L be the line at which they intersect. Find the point at which L intersects the xy-plane. The point being on the line at the intersection of the two planes must satisfy the equation of both planes. In addition, the point is in the xy-plane, so it must satisfy z 0. Thus, we need to solve x + y z 1 2x y + 4z 5 z 0 That is { x + y 1 2x y 5 From the first equation, we get. Replacing in the second equation gives Replacing in equation 1 gives y 1 x (1) 2x (1 x) 5 2x + x 5 x 8 5 y 1 8 5 5 Thus the coordinates of the point in question are ( 8 5, ) 5, 0 6. Find the equation of the plane through ( 4, 1, 2) parallel to the plane x + 2y + 5z (7 points) The plane we want the equation of and the given plane have the same normal since they are parallel. This normal is 1, 2, 5. The equation of the plane through (x 1, y 1, z 1 ) with normal a, b, c is In our case, we obtain a (x x 1 ) + b (y y 1 ) + c (z z 1 ) 0 (x + 4) + 2 (y 1) + 5 (z 2) 0 x + 2y + 5z 8 7. Find the velocity vector v (t) and the position r (t) of a particle if its acceleration is a (t) 0, 0, 1, its initial velocity is v (0) 1, 1, 0 and its initial position is r (0) 0, 0, 0. (7 points) 4
The velocity is obtained by integrating the acceleration. integrated by integrating each component. Thus, v (t) 0dt, 0dt, dt Vector functions are C 1, C 2, t + C Thus, on one hand v (0) C1, C 2, C On the other hand, we are given that v (0) 1, 1, 0. Therefore, we must have: C 1, C 2, C 1, 1, 0 It follows that C 1 1, C 2 1 and C 0. Thus, v (t) 1, 1, t The position is obtained by integrating the velocity. Thus, r (t) dt, ( 1) dt, tdt t + C 1, t + C 2, 12 t2 + C On one hand r (0) C1, C 2, C On the other hand, we are given that r (0) 0, 0, 0. Therefore, we must have: C 1, C 2, C 0, 0, 0 So, we see that the three constants are 0. Thus, r (t) t, t, 1 2 t2 8. Find the parametric equations of the tangent line to the curve C with position vector r (t) 2 t, 2t 1, ln t at the point (1, 1, 0). (7 points) The given point corresponds to t 1. Since we are given a point, we only need a direction vector. Such a vector is r (1). Since r (t) t 2, 2, 1 t, it follows that r (1), 2, 1. Therefore, the parametric equations of the line are x 1 t y 1 + 2t z t 5