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Unformatted text preview: MATH 20C — MIDTERM #1 REVIEW SHEET SOLUTIONS 1. Section 12.1, problems 21 and 22. A sketch for problem 21 is in the back of the book. 2. Consider the 12 vectors which have their tails at the center of a clock and their respective heads at each of the 12 digits. What is the sum of these vectors? Each of the 12 vectors can be paired with the one diametrically opposite (the vector pointing to 3 with that pointing to 9, the one pointing to 4 with that pointing to 10, etc.), which is its negative. Since v + ( v ) = , the sum of all 12 vectors is the zero vector. 3. What are the components of a vector in two dimensions which has length 5 and is directed halfway between the positive yaxis and the negative xaxis? The indicated direction makes an angle of 3 π 4 with the positive xaxis. Hence its coordinates are h 5cos 3 π 4 , 5sin 3 π 4 i = h 5 √ 2 , 5 √ 2 i . 4. Give (in components) two unit vectors which lie on the line 3 y = 4 x . The vector h 3 , 4 i lies on the given line, since 3 · 4 = 4 · 3. Its magnitude is kh 3 , 4 ik = √ 3 2 + 4 2 = 5. Thus, a unit vector pointing in the same direction as h 3 , 4 i is 1 5 h 3 , 4 i = h 3 5 , 4 5 i . One pointing in the opposite direction is h 3 5 , 4 5 i . 5. A sailboat is sailing at an angle of 60 ◦ east of south. A wind blowing from due north exerts a force of 150 Newtons on the sail, directed due south. What magnitude of force pushes the sailboat in the direction it wants to go? We are looking for the projection of the wind force vector onto the direction of the sailboat’s motion. Thus, let F be the vector of the wind, and let e be a unit vector in the latter direction. The angle between F and e is given as 60 ◦ , so F · e = k F k · k e k · cos60 ◦ = 150 · 1 · 1 2 = 75 Newtons. 6. Section 12.1, problems 59 and 60. For problem 59, we have F 2 = h 1 √ 2 f 2 , 1 √ 2 f 2 i (where f 2 is its magnitude), and F 2 = h ,f 1 i , while the third force is h 10 √ 3 , 10 i . Thus if the net force is zero, we have 1 √ 2 f 2 = 10 √ 3 f 1 1 √ 2 f 2 = 10 from which we conclude f 2 = 10 √ 6 and so f 1 = 10(1 + √ 3). For problem 60, we have v 1 = h 200 , i and v 2 = h 20 √ 2 , 20 √ 2 i , so that v = h 200 + 20 √ 2 , 20 √ 2 i ≈ h 228 . 3 , 28 . 3 i and k v k ≈ 230. 7. Calculate 4 h 3 , 5 , 1 i + 8 h 1 , 2 , i . h 4 · 3 + 8 · ( 1) , 4 · 5 + 8 · 2 , 4 · ( 1) + 8 · i = h 4 , 36 , 4 i . 8. Find a parametric equation of the line in 3dimensional space which is perpen dicular to the xyplane and passes through the point (2 , 1 , 4) . A vector perpendicular to the xyplane is k = h , , 1 i . We know that the general equation of a line with direction vector v passing through a point with direction 1 2 MATH 20C — MIDTERM #1 REVIEW SHEET SOLUTIONS vector r is r ( t ) = r + t v . Thus, here we have r ( t ) = h 2 , 1 , 4 i + t h , , 1 i = h 2 , 1 , 4 + t i ....
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This note was uploaded on 12/08/2011 for the course MATH 20C 20C taught by Professor Boonyeap during the Fall '07 term at UCSD.
 Fall '07
 BoonYeap

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