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Unformatted text preview: Math 152, Spring 2011 Solutions Assignment #5 Notes: • Each question is worth 5 marks. • Due in class: Monday, February 7 for MWF sections; Tuesday, Febru ary 8 for TTh sections. • Solutions will be posted Tuesday, February 8 in the afternoon. • No late assignments will be accepted. (1) (a) (1 mark) Find the equation form of the plane P with normal vector b = [1 , , 1] and passing through the point q = [0 , 1 , 0]. (b) (1 mark) Find the equation form of the line L that is perpendicular to the plane Q with equation x +3 y +2 z = 1 and passing through the origin. (c) (3 marks) Use Gaussian elimination to determine the intersection between the plane P and the the line L . Solution: (a) The equation form of the plane P is x 1 + 0 x 2 x 3 = b · q = 0; that is x 1 x 3 = 0 . (b) Since the equation form of the plane Q is the equation x 1 + 3 x 2 + 2 x 3 = 1 , then a normal vector to the plane Q is the vector b 1 = [ 1 , 3 , 2]. This vector is in turn parallel to the line L . Since the line L passes through the origin its parametric equations are x = t b 1 = t [ 1 , 3 , 2]; that is, x 1 = t, x 2 = 3 t, x 3 = 2 t. We can use these to find the equation form of the line L . In this case we obtain 3 x 1 + x 2 = 0 2 x 1 + x 3 = 0 . (c). The intersection of the plane P and the line L is the solution of the system of equation x 1 x 3 = 0 3 x 1 + x 2 = 0 2 x 1 + x 3 = 0 . The augmented matrix associated to this system is 1 2 1 0 1  3 1  2 0 1  Applying Gaussian elimination we obtain the echelon form 1 0 1  0 1 3  0 0 3  We can use this system using backwards substitution. In this case we get x 1 = x 2 = x 3 = 0. Thus the intersection between P and L is the point [0 , , 0]. (2) (a) (2 marks) Use Gaussian elimination to find the set of solutions to the following homogeneous system. Present your answer in parametric form....
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This note was uploaded on 03/30/2011 for the course MATH 152 taught by Professor Caddmen during the Spring '08 term at UBC.
 Spring '08
 Caddmen
 Math

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