HOMEWORK 1 SOLUTIONS. It s helpful to write all vectors in component form, then subtract componentwise:

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1 HOMEWORK 1 SOLUTIONS Section 1.1 Problem 4 It s helpful to write all vectors in component form, then subtract componentwise: Problem 10 (2, 3, 5) (4, 0, 0) + (0, 3, 0) = ( 2, 6, 5). I ll omit the sketch here, and won t ask you to sketch 3-dimensional vectors on an exam. The vectors point in opposite directions because the first vector is -3 times the second vector. Being scalar multiples of each other indicates that they re parallel, and the negative sign indicates that they point in opposite directions. Problem 11 To lie on the y-axis, x and z must equal 0. To lie on the z-axis, x and y must be 0. To lie in the xz plane, y must be 0. To lie in the yz plane, x must be 0. Problem 16 The vector equation of the given line is l(t) = (0, 2, 1) + t(2, 0, 1). We first find a vector parallel to the line by subtracting the two points we know to be on the line: ( 5, 0, 4) (6, 3, 2) = ( 11, 3, 2). Now we can write the vector equation of the line as l(t) = ( 5, 0, 4) + t( 11, 3, 2). Note: This equation is not unique. If you had subtracted the points in the reverse order your equation would look different but still represent the same line. Think about why! To show two equations parameterize the same line we need to show that their direction vectors are parallel (so the lines are parallel) and that the lines have a point in common (so that they actually coincide). We have ( 2, 0, 4) = 2(1, 0, 2) so the lines are parallel. Setting t = 1 in the first equation and t = 0 in the second shows that both lines contain the point (2, 2, 1). Therefore the lines are the same. 1

2 2 HOMEWORK 1 SOLUTIONS Problem 24 For the xy plane, we set z = 0, so that t = 2. Substituting this in, we see that the line intersects the xy plane at (7, 23, 0). Similarly for the xz plane we set y = 0 so t = 7/8. The point of intersection is (5/4, 0, 23/8). For the yz plane, set x = 0. The point of intersection is (0, 5, 7/2). Problem 26 A typical point on the line is of the form (1 + 2t, 1 + 3t, 2 + t). Plugging this into the equation we have: Problem 28 5x 3y z 6 = 5(1 + 2t) 3( 1 + 3t) (2 + t) 6 = t + 3 9t 2 t 6 = 0 For the lines to intersect, we need to find t and s such that (t + 4, 4t + 5, t 2) = (2s + 3, s + 1, 2s 3). Comparing these componentwise, we must have: t + 4 = 2s + 3 4t + 5 = s + 1 t 2 = 2s 3 From the first equation we have t = 2s 1. Plugging this into the second equation we have 8s = s + 1 so that s = 0. This implies t = 1. Since these values also satisfy the third equation we conclude that the system has a solution and the lines intersect. Section 1.2 Problem 6 u = 245 v = π u v = 15π 10. Problem 12 Let w = (a, b). Since w = 5 we must have a 2 + b 2 = 25. Since w and v are orthogonal we must have 2a + 3b = 0. Solving these two equations, one possible solution is ( 15/ 13, 10/ 13). Problem 15 We know v w = v w cos(θ). If this is equal to v w then we must have cos(θ) = 1, so θ = π. This means the vectors point in opposite directions. Setting the dot product equal to zero means x 2 + x 6 = 0. Solutions to this are x = 2 and x = 3.

3 HOMEWORK 1 SOLUTIONS 3 Using the formula for projection we have v u v 2 v = 4 (2, 1, 3). 14 Problem 23 We know v w = v w cos(θ) = cos(θ). Since the angles of an equilateral triangle are all π/3, the dot product is 1/2. Problem 30 (a) With no current, the ship s velocity relative to the bottom is represented by (0, 4). (b) If the ship were just drifting, its velocity relative to the bottom is represented by (1, 0). (c) The total velocity of the ship is the sum of these vectors: (1, 4). (d) After 1 hour, the ship would be at position (1, 0) + (1, 4) = (2, 4). This coincides with where the rock is! (e) The captain should change course if he intends to sail for more than an hour. (f) If the rock were an iceberg, this problem would be an oversimplified version of why the Titanic sunk. Section 1.3 Problem 4 First we compute the cross product: b c = (1 2 1 ( 1))i ( )j + (( 1) 2 1 3)k = 3i j 5k. Taking the dot product we have Problem 5 a (b c) = (1, 2, 1) (3, 1, 5) = = 0. To find the area of the parallelogram, take the cross product of a and b, then compute its magnitude. The result is 35. Problem 6 This problem is just like example 8 of section 1.3. We use the fact that the area of the triangle is half the area of the parallelogram spanned by two sides of the triangle. Subtracting, we have that (1, 1, 1) and (0, 2, 3) are two sides of the triangle, so we ll just compute the magnitude of their cross product: Problem 8 (1, 1, 1) (0, 2, 3) /2 = (5, 3, 2) /2 = 38/2. We need to compute the absolute value of the triple product of the three vectors: ((1, 0, 0) (0, 3, 1)) (4, 2, 1) = (0, 1, 3) (4, 2, 1) = 1.

4 4 HOMEWORK 1 SOLUTIONS Problem 10 First we compute the cross product to find a single orthogonal vector: ( 5, 9, 4) (7, 8, 9) = (113, 17, 103). Now we normalize this vector to find a unit vector pointing in the same direction, and note that the unit vector pointing in the opposite direction also satisfies the conditions of the problem. (113, 17, 103) = so the unit vectors orthogonal to the given vectors are ± 1 7 (113, 17, 103). 483 Problem 15 (a) x + y + z = 1 (b) x + 2y + 3z = 6 (c) 5x + 2z = 25 (d) x 2y + 3z = 13 Problem 16 (a) 4x + 6y + 8z = 0 (b) 5x 5y + 5z = 15 (c) 20x + 2y + 19z = 95 Points (b) and (c) are contained in P because they satisfy the equations x + y + z = 1. Two planes intersect along a line, which is determined by a point and a direction vector. Since this line lies in both planes, it is orthogonal to the normal vectors of both planes. To find the direction vector of this line, we compute the cross product of the normal vectors: (1, 2, 1) (1, 3, 1) = (1, 2, 5). Now we need to find a point on the intersection. It s easy to see that both planes go through the origin, so we ll use (0, 0, 0). The equation of the line is Problem 26 l(t) = (t, 2t, 5t). We know v w = v w sin(θ), so if this also equals 1 2 v w then we must have sin(θ) = 1/2, so the angle between the vectors is either π/6 or 5π/6. Problem 31 The vector (2, 3, 1) is parallel to the plane, but we need another one. We can find this by subtracting the two points we re given on the lines: (2, 1, 0) (0, 1, 2) = (2, 2, 2). Now we compute the cross product to find a normal vector:

5 (2, 3, 1) (2, 2, 2) = (4, 6, 10). The plane equation is 4x 6y 10z = 14. HOMEWORK 1 SOLUTIONS 5