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Course: M 408D 408D, Spring 2012
School: University of Texas
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gilbert DiscGQ06 (55035) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 Which of the following expressions are well-defined for all vectors a, b, c, and d? 10.0 points Find an equation for the plane passing through the origin and parallel to the plane (a b) (c d) , II 001 I a (b c) , III a (b c) ....

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gilbert DiscGQ06 (55035) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 Which of the following expressions are well-defined for all vectors a, b, c, and d? 10.0 points Find an equation for the plane passing through the origin and parallel to the plane (a b) (c d) , II 001 I a (b c) , III a (b c) . 1. I and III only x + 4 y 2z = 1 . 2. III only 1. 4x + 2y + 4z = 0 3. II only 2 . 4 x 2y 4z + 1 = 0 4. I and II only 3 . 2x + y 4z + 1 = 0 5. II and III only correct 4. x + 4y 2z = 0 correct 6. I only 5 . 2 x y + 4z = 0 7. all of them 6. x 4 y + 2z + 1 = 0 8. none of them Explanation: The scalar equation for the plane through P (a, b, c) with normal vector n = Ai + Bj + Ck is A(x a) + B(y b) + C(z c) = 0 . But P (a, b, c) = (0, 0, 0) if the plane passes through the origin, while n = i + 4 j 2k if the plane is parallel to x + 4 y 2z = 1 since parallel planes have the same normal vector. Consequently, the plane has equation x + 4 y 2z = 0 . Explanation: The cross product is defined only for two vectors, and its value is a vector; on the other hand, the dot product is defined only for two vectors, and its value is a scalar. For the three given expressions, therefore, we see that I is not well-defined because each term in the cross product is a dot product, hence a scalar. II is well-defined because it is the dot prod-uct of two vectors. III is well-defined because it is the cross product of two vectors. keywords: vectors, dot product, cross prod-uct, T/F, length, 003 10.0 all points Determine unit vectors v orthogonal to 002 10.0 points a = i + 3 j + 4k, b = 3 i + 12 j + 14 k . DiscGQ06 gilbert (55035) 3 1. v = 7i+ 2 7j+ 6 4 7k 3. v = i 7 2 7 j 10.0 points A line passes through the point P (4, 2, 2) and is parallel to the vector h 1, 4, 2 O. 2. v = 3 i + 2 j + 6 k 3 2 6 7 k At what point Q does intersect the xy-plane? 3 1. Q(0, 3, 6) 4. v = 7 i 7 j + 7 k 2. Q(5, 6, 0) 5. v = 6 i 2 j + 3 k 3. Q(0, 6, 5) 6 2 6 6. v = 2 i+ 7 7 j 3 7k 4. Q(0, 5, 2) correct Explanation: The non-zero vectors orthogonal to a and b are all of the form v = (a b) , v | a b| . = 6. Q(3, 2, 0) correct Explanation: Since the xy-plane is given by z = 0, we have to find an equation for and then set z = 0. Now a line passing through a point P (a, b, c) and having direction vector v is given parametrically by =6 0 , with a scalar. The only unit vectors orthog-onal to a, b are thus ab 5. Q(2, 3, 0) r(t) = a + tv , But for the given vectors a and b, i ab = =3 12 k j 1 3 3 4 But for , 14 12 a = h 4, 2, 2 O , i 1 4 j + 1 4 14 3 a = h a, b, c O . 14 3 3 k 12 = 6 i 2 j + 3 k . v = h 1, 4, 2 O . Thus r(t) = h 4 + t, 2 + 4t, 2 + 2t O , so z = 0 when t = 1. Consequently, the line intersects the xy-plane at In this case, Q(3, 2, 0) 2 |a b| = 49 . . Consequently, v= 6 2 3 7i+ 7j 7 k keywords: vector product, cross product, unit vector, orthogonal, . keywords: line, parametric equations, direc-tion vector, point on line, intercept, coordi-nate plane
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