MATH Calculus III Fall 2009 Homework 1 - Solutions

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1 MATH Calculus III Fall 2009 Homework 1 - Solutions 1. Find the equations of the two spheres that are tangent with equal radii whose centers are ( 3, 1, 2) and (5, 3, 6). SOLUTION: In order for the spheres to be tangent with equal radii, the radius must be half of the distance between the two points: r 1 (5 ( 3))2 + ( 3 1) (6 2) Therefore, our two spheres will be: (x + 3) 2 + (y 1) 2 + (z 2) 2 24 (x 5) 2 + (y + 3) 2 + (z 6) Let a i 2j+3k, b 4i j+2k, and c 2i 3k. Find the value of x such that a+xb is orthogonal to c. SOLUTION: First, we find the vector a + xb (1 + 4x)i (2 + x)j + (3 + 2x)k. In order to be orthogonal to c, we know that (a + xb) c 0. Therefore, we have: (a + xb) c ((1 + 4x)i (2 + x)j + (3 + 2x)k) (2i 3k) 2 + 8x 9 6x 2x 7 0 Therefore, x Find a vector of length 3 which is orthogonal to both u 2i + 3j 4k and v i 5j k. SOLUTION: First, we find a vector which is orthogonal to both u and v by finding the cross product: u v i 2j 13k Then, we construct a vector of length 3 by finding a unit vector parallel to u v and multiplying it by a scaling factor of 3: ( ) ( ) u v 23i 2j 13k i 2 j 13 k u v Note that minus the answer obtained is also correct. For problems 4 and 5, use the points A(3, 2, 1), B( 1, 0, 7), C(4, 4, 5), and D(1, 3, 2). 4. Find the following: (a) AB ( BC AD) SOLUTION: AB ( BC AD) 4, 2,6 ( 5,4, 2 2,1, 3 ) 56( ) 0

2 MATH 2300 Fall 09 HW 1 Solutions 2 (b) proj BD ( AC + AB) SOLUTION: proj BD ( AC + AB) ( AC + AB) BD BD BD 2 (c) the angle (in radians) between DA and BC ( 1,2,4 + 4, 2,6 ) 2,3, 9 2,3, , , ,3, 9 94 SOLUTION: ( θ cos 1 DA BC ) DA BC ( ) 2, 1,3 5,4, 2 cos 1 cos 1 (0) π Find the following lines and plane: (a) the parametric equations of the line containing D and parallel to the y-axis SOLUTION: The slope vector for the y-axis is the vector v j, and we can use the same slope vector for the parallel line we wish to construct. Then, using the point D, we have: r r 0 + tv 1,3, 2 + t 0,1,0 1,3 + t, 2 Then the parametric equations for the line are: x 1,y 3 + t,z 2. (Answers may vary.) (b) the symmetric equations of the line passing through A and orthogonal to the plane containing B, C, and D SOLUTION: First, we find the normal vector to the plane containing B, C, and D by finding the cross product: BC CD ,41,7 This vector will serve as the slope vector of the line we will construct. Then, we begin with the vector equation of the line: r r 0 + tv 3,2,1 + t 30,41,7 3 30t,2 + 41t,1 + 7t The parametric equations are x 3 30t,y t,z 1 + 7t, and we find the symmetric equations by solving for t: (c) the equation of the plane containing A, B, and C x 3 30 y 2 z SOLUTION: We begin by finding the normal vector to the plane from the cross product: n AB BC ,22, Then, using the vector equation of the plane, we use any of the three points for r 0, and we have: n (r r 0 ) 20,22, 6 x 3,y 2,z 1 20x y 44 6z The simplified equation of the plane is then 10x 11y + 3z 11

3 MATH 2300 Fall 09 HW 1 Solutions 3 6. (a) Let a a 1, a 2, a 3 and b b 1, b 2, b 3. Show that (a + b) (a b) a 2 b 2. SOLUTION: (a + b) (a b) a (a b) + b (a b) Property (3) p. 815 (a a) (a b) + (b a) (b b) a 2 (a b) + (b a) b 2 Property (3) p. 815 a 2 (a b) + (a b) b 2 Property (1) p. 815 a 2 b 2 Property (2) p. 815 (b) Show that for any two nonzero vectors u and v, the vectors x v u + u v and y v u u v are orthogonal. SOLUTION: METHOD 1 - In order to show that x and y are orthogonal, we must show that their dot product will always equal 0. We note that this is a particular case of part (a) with a v u and b u v. Notice here that a 2 b 2, so we have x and y are orthogonal. SOLUTION: METHOD 2 - In order to show that x and y are orthogonal, we must show that their dot product will always equal 0. ( v u + u v) ( v u u v) v u ( v u u v) + u v ( v u u v) Property (3) p. 815 ( v u v u) ( v u u v) + ( u v v u) ( u v u v) Property (3) p. 815 v 2 (u u) v u (u v) + u v (v u) u 2 (v v) Property (4) p. 815 v 2 u 2 v u (u v) + u v (v u) u 2 v 2 Property (1) p. 815 v 2 u v u (u v) + u v (u v) u 2 v Property (2) p. 815 v u (u v) + u v (u v) 0 SOLUTION: METHOD 3 - Let a v u, and b u v. Then, our task is to show that x y, or that a +b a b. First, note that a and b have the same length: Therefore, we can sketch a, b, x, and y as shown below: a v u v u b u v u v Since x and y form the diagonals of the rhombus (parallelogram with sides of equal length) formed by a and b, we recognize that the two vectors (diagonals) x and y are orthogonal. 7. (a) Find parametric equations for the line of intersection between the planes x + y 0 and 3y + 8z 6. SOLUTION: First we find a vector that is parallel to the line of intersection by finding the cross product of the normal vectors: v n 1 n , 8,

4 MATH 2300 Fall 09 HW 1 Solutions 4 Then, we find a point which is on the line of intersection by selecting one value, and solving for the other two coordinates. ( For example, we could use y 0, which would give us the two equations x 0 and 8z 6, or a point of 0,0, 3 ). (Note that the choice of the point here is not unique, so answers may vary.) 4 Now, using the vector and the point to construct the equation of the line, we have: r r 0 + tv 0,0, t 8, 8,3 8t, 8t, t Thus, the parametric equations of the line of intersection are: x 8t,y 8t,z t. (b) Find the point of intersection of the line found in part (a) and the plane 2x + 5y + 4z 5. SOLUTION: To find the point of intersection of the line and the plane, we substitute the values from the parametric equations in part (a) into the equation of the plane: ( ) 3 2(8t) + 5( 8t) t 5 16t 40t t 5 12t 2 Then, we have t 1 ( 6, and the point of intersection is 4 3, 4 3, 1 ) Given the intersecting lines: L 1 : x 3t 3, y 2t, z 6t + 7 L 2 : x s 6, y 3s 5, z 2s + 1 (a) Find the (acute) angle, rounded to the nearest degree, between the lines. SOLUTION: First, we find the slope vectors of the lines, v 1 3, 2,6, and v 2 1, 3,2. Then the angle between the lines is the same as the angle between their slope vectors: ( ) ( ) ( ) θ cos 1 v1 v 2 3, 2,6 1, 3,2 3 cos 1 cos 1 37 v 1 v (b) Find the point of intersection of the two lines. SOLUTION: We find the point of intersection by solving the system of equations: 3t 3 s 6 2t 3s 5 s 3, t 2 6t + 7 2s + 1 Plugging these values into our lines gives us the point of intersection ( 9,4, 5) (c) Find the equation of the plane containing the two lines. SOLUTION: We find the normal vector to the plane by finding the cross product of the slope vectors of the lines: n v 1 v ,0, Then we use the point of intersection to find the equation of the plane: n (r r 0 ) 14,0, 7 x + 9,y 4,z x z 35 0 The simplified answer becomes 2x z 13.

5 MATH 2300 Fall 09 HW 1 Solutions 5 9. Given the lines: L 1 : x 1 2t, y 6t, z 8t L 2 : x 3 + t, y 3t, z 5 4t (a) Show that the lines are parallel. SOLUTION: We can demonstrate that the two lines are parallel by showing that their slope vectors are parallel. Using v 1 2, 6,8, and v 2 1,3, 4, we can do this in two ways. First, we can recognize that v 1 2v 2. Since the vectors are scalar multiples of one another, they are parallel. Second, we can calculate v 1 v 2 0. Since the cross product is the zero vector, the two vectors are parallel, and hence the lines are parallel. (b) Find the distance between the two lines. SOLUTION: METHOD 1 - To find the distance between the two lines, we will chose a point on L 1, and find its distance to the line L 2. This distance will be uniform regardless of the point we choose, since the lines are parallel. On L 1, we ll use t 0 to determine the point P( 1,0,0). Similarly, we will also choose a point on L 2, by setting t 0, which gives us Q(3,0,5). Now, let x QP 4,0, 5. We will use the orthogonal projection, orth v x x proj v x, whose length will be the distance between the two lines. (The vector v is the slope vector of line L 2, or v 1,3, 4.) [See p. 821 #41 for more information on the orthogonal projection] d orth v x x proj v x ( ) 4,0, 5 1,3, 4 4,0, 5 1,3, 4 2 1,3, 4 8 4,0, 5 1,3, , 24 13, SOLUTION: METHOD 2 - This problem can also be solved by finding t that minimizes the distance from the point P ( 1,0,0) to the point T (3 + t,3t,5 4t). The distance between P and T is given by the length of TP, ( 4 t) 2 + ( 3t) 2 + ( 5 + 4t) 2. Let the square of the distance between these two points be g(t) ( 4 t) 2 + ( 3t) 2 + ( 5 + 4t) 2 26t 2 32t Now we must find t such that g (t) 0. We find g (t) 52t 32 and therefore when t 8/13, g (t) 0 and a minimum value occurs. Therefore, the distance between the two lines is given by 405 g(8/13) 13. (c) Find the equation of the plane containing the two lines. SOLUTION: We can find the equation of the plane containing the parallel lines by first finding the normal vector. We can do this by finding the cross product of the vectors x and v used in part (b): n x v , 21, Then, using the point P, the equation of the plane is: n (r r 0 ) 15, 21, 12 x + 1,y,z 15x y 12z 0 The simplified answer becomes 5x 7y 4z Is the line x 1 + 2t, y 2 + 3t, z 5t orthogonal to the plane 4x 6y + 10z 9? Give reasons to justify your answer.

6 MATH 2300 Fall 09 HW 1 Solutions 6 SOLUTION: First, we examine the slope vector of the line, v 2,3, 5, and the normal vector of the plane, n 4, 6,10. Notice that n 2v, which indicates that the two vectors are parallel. Since the line is parallel to the normal vector of the plane, the relationship is that the line is orthogonal to the plane. 11. Consider the points P 1 (sin2, 3, 1), P 2 (5, 3.7, π), P 3 (e + 1, 11.2, 9), and P 4 (1, ln 2, 6). Use a computer to answer the following questions. Print your input and output for submission with HW1. (a) Define the vectors U P 1 P 2, V P 1 P3, and W P 1 P 4. Find the volume (vol1) of the parallelepiped formed by the three vectors. (b) Consider three additional vectors, which lie along the diagonals of the parallelpiped on the faces adjacent to P 1. Define these vectors, calling them X, Y, and Z. Find the volume (vol2) of the second parallelpiped formed by these vectors. What is the ratio of vol2/vol1? (c) Find the angle, in radians, between the main diagonals of the parallelepipeds that contain the point P 1. SOLUTION: In Mathematica, the commands might be: p1{sin[2],3,-1}; p2{5,-3.7,pi}; p3{e+1,11.2,9}; p4{1,log[2],-6}; Up2-p1; Vp3-p1; Wp4-p1; vol1abs[u.(v W)] with output: vol Further, for part (b), the Mathematica commands might be: XU+V; YU+W; ZV+W; Vol2Abs[X.(Y Z)] with output: vol , vol2/vol1 with output: 2 and (U+V +W).(X+Y +Z) ArcCos[ ] with output: 0. (U+V +W).(U+V +W) (X+Y +Z).(X+Y +Z) Thus, the ratio is 2 so the second parallelepiped has twice the volume of the first. Also, the angle between the main diagonals is found to be 0, therefore the diagonals are parallel. The two diagonals must also be collinear since they have a point P 1 in common.