2.2. HW #2: Due Wednesday, February 6, 2013: Solutions: Page 823: #9:: Find | −→a |, | − 2 −→b |, | −→a − −→b |, −→a + −→b and 3 −→a − 2 −→b for −→a = h1, −2i and −→b = h−3, 2i. Solution: We have | −→a | = p 1 2 + (−2)2 = √ 5. As −2 −→b = h6, −4i, | − 2 −→b | = p 6 2 + (−4)2 = √ 52. Since −→a − −→b = h4, −4i, | −→a − −→b | = p 4 2 + (−4)2 = √ 32. Finally, −→a + −→b = h−2, 0i and 3 −→a − 2 −→b = h3, −6i − h−6, 4i = h9, −10i. Page 823: #18:: Find a unit vector −→u in the same direction as −→a = h5, −12i. Express −→u in terms of −→i and −→j , and find a vector −→v in the opposite direction as that of −→a . Solution: We have | −→a | = p 5 2 + (−12)2 = √ 169 = 13. A unit vector is −→u = −→a /| −→a |, or −→u = h5/13, −12/13i. As −→i = h1, 0i and −→j = h0, 1i, we have −→u = 5 13 −→i − 12 13 −→j . As − −→a has the opposite direction as −→a , we see we may take −→v = − −→a = h−5, 12i. Of course, there are multiple answers. We could also take −→v = − −→u , as −→u and −→a are in the same direction. Page 823: #38:: Given three points A(2, 3), B(−5, 7) and C(1, −5), verify by direct computation that −−→AB + −−→BC + −→CA is the zero vector. Solution: Given two points P = (p1, p2) and Q = (q1, q2), by −−→P Q we mean the vector from P to Q, which is hq1 − p1, q2 − p2i. We thus have −−→AB = h−5, 7i − h2, 3i = h−7, 4i −−→BC = h1, −5i − h−5, 7i = h6, −12i −→CA = h2, 3i − h1, −5i = h1, 8i, which implies −−→AB + −−→BC + −→CA = h−7, 4i + h6, −12i + h1, 8i = h0, 0i. Why is this true? We are traveling in a directed way along the three edges of a triangle, and we return to where we started. Page 824: #42:: Let −→a = ha1, a2i and −→b = hb1, b2i. Prove by componentwise arguments that if −→a + −→b = −→a then −→b = −→0 . Solution: Assume −→a + −→b = −→a . Substituting for these vectors yields ha1, a2i + hb1, b2i = ha1, a2i, or equivalently ha1 + b1, a2 + b2i = ha1, a2i. This is a pair of equations: a1 + b1 = a1, a2 + b2 = a2. We now have simple equations of numbers and not vectors. For the first, subtracting a1 from both sides gives b1 = 0, while for the second subtracting a2 from both sides gives b2 = 0. Thus our vector −→b = h0, 0i. The key observation here is that we can reduce a vector question to a system of equations about numbers, and we know how to handle / analyze numbers. Page 833: #1:: Let −→a = h2, 5, −4i and −→b = h1, −2, −3i. Find 2 −→a + −→b , 3 −→a − 4 −→b , −→a · −→b , | −→a − −→b | and −→a /| −→a |. Solution: First, 2 −→a + −→b = h4, 10, −8i + h1, −2, −3i = h5, 8, −11i 3 −→a − 4 −→b = h6, 15, −12i − h4, −8, −12i = h2, 23, 0i.