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**Unformatted text preview: **MATH 51 MIDTERM 1 SOLUTIONS October 16, 2008 1. Find all solutions of the following system: x 1 + 2 x 2 + x 3 + x 4 = 7 x 1 + 2 x 2 + 2 x 3- x 4 = 12 2 x 1 + 4 x 2 6 x 4 = 4 . Solution : See page 44 of the text. 2(a). Find a parametric equation for the plane containing the points A = (1 , 2 , 3), B = (4 , 5 , 6), and C = (2 , 2 , 3). 2(b). Find the equation for the plane that passes through the point A = (1 , 2 , 3) and that is perpendicular to the vector v = 7 3 5 . (Your answer should be an equation of the form ax + by + cz = d .) Solution : Note that P = x y z is in the plane if and only if v is orthogonal to- A P , i.e., if and only if v - A P = 0, i.e., if and only if v ( P- A ) = 0 . Thus the equation is 7 3 5 x y z - 1 2 3 = 0 or, equivalently, 7 x + 3 y + 5 z = 28 . 3(a) Suppose that is an equilateral triangle in R 3 and that the edges of each have length 1. Let A , B , and C be the vertices of . Find (3- AB ) (5- AC ) . Solution : (3- AB ) (5- AC ) = 15- AB - AC = 15 k- AB kk- AC k cos = 15 1 1cos 2 = 15 2 . 1 3(b) . Suppose A = a 1 a 2 a 3 a 4 and B = b 1 b 2 b 3 b 4 are orthogonal vectors in R 4 with a 4 > 0 and b 4 > 0. Let a = a 1 a 2 a 3 and b = b 1 b 2 b 3 . Prove that the angle between a and b is obtuse (i.e., greater than / 2). Solution : 0 = A B = a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = a b + a 4 b 4 > a b since a 4 and b 4 are positive. Thus (*) > a b = k a kk b k cos (where is the angle between a and b ). Since k a k and k b k are nonnegative, (*)...

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