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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 Δ is an equilateral triangle in R 3 and that the edges of Δ each have length 1. Let A , B , and C be the vertices of Δ. Find (3→ AB ) · (5→ AC ) . Solution : (3→ AB ) · (5→ AC ) = 15→ AB ·→ AC = 15 k→ AB kk→ AC k cos θ = 15 · 1 · 1cos ‡ π 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 π/ 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 θ (where θ is the angle between a and b ). Since k a k and k b k are nonnegative, (*)...
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 Math, Linear Algebra, Algebra, Differential Calculus, 5w, 0 2 1 1 hand

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