soln4 - SOLUTIONS TO ASSIGNMENT #4 Math 152 1. Consider the...

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Unformatted text preview: SOLUTIONS TO ASSIGNMENT #4 Math 152 1. Consider the intersection between the following two planes given in parametric form: P 1 : x = [2 , 4 , 3] + s 1 [1 , 2 , 1] + s 2 [2 , 5 , 4] P 2 : x = [1 , ,- 5] + t 1 [3 , 8 , 7] + t 2 [2 , 1 ,- 5] Find the intersection as a line in parametric form. Solution: We will solve the equation x = [2 , 4 , 3] + s 1 [1 , 2 , 1] + s 2 [2 , 5 , 4] = [1 , ,- 5] + t 1 [3 , 8 , 7] + t 2 [2 , 1 ,- 5] or s 1 [1 , 2 , 1] + s 2 [2 , 5 , 4]- t 1 [3 , 8 , 7]- t 2 [2 , 1 ,- 5] = [- 1 ,- 4 ,- 8] for s 1 , s 2 , t 1 and t 2 . To solve, we write the system as an augmented matrix and reduce it to row-echelon form. 1 2- 3- 2- 1 2 5- 8- 1- 4 1 4- 7 5- 8 1 2- 3- 2- 1 0 1- 2 3- 2 0 2- 4 7- 7 R 2 R 2- 2 R 1 R 3 R 3- R 1 1 2- 3- 2- 1 0 1- 2 3- 2 0 0 1- 3 R 3 R 3- 2 R 2 Therefore, t 1 (the 3rd variable) is a free variable and by back substitution, t 2 =- 3 s 2- 2 t 1 + 3 t 2 =- 2 s 2 = 7 + 2 t 1 s 1 + 2 s 2- 3 t 1- 2 t 2 =- 1 s 1 =- 21- t 1 We actually do not need s 1 and s 2 because we can just plug t 2 =- 3 into P 2 and get the equation of the line x = [1 , ,- 5] + t 1 [3 , 8 , 7]- 3[2 , 1 ,- 5] or x = [- 5 ,- 3 , 10] + t 1 [3 , 8 , 7] 2. Reduce the matrix - 1 2 0 1 4 1- 2 0 0- 1 2- 6 2 4 to reduced row echelon form and determine the rank of the matrix. Solution: - 1 2 0 1 4 1- 2 0 0- 1 2- 6 2 4 - 1 2 0 1 4 0 1 3- 2 2 6 8 R 2 R 2 + R 1 R 3 R 3 + 2 R 1 - 1 2 0 1 4- 2 2 6 8 0 1 3 R 2 R 3...
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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 The University of British Columbia.

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soln4 - SOLUTIONS TO ASSIGNMENT #4 Math 152 1. Consider the...

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