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soln4

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

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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 , 0 , - 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 , 0 , - 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 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 , 0 , - 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 0 to reduced row echelon form and determine the rank of the matrix.

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Solution: - 1 2 0 1 4 1 - 2 0 0 - 1 2 - 6 2 4 0 - 1 2 0 1 4 0 0 0 1 3 0 - 2 2 6 8 R 2 R 2 + R 1 R 3 R 3 + 2 R 1 - 1 2 0 1 4 0 - 2 2 6 8 0 0 0 1 3 R 2 R 3 R 3 R 2 - 1 2 0 0 1 0 - 2 2 0 - 10 0 0 0 1 3 R 1 R 1 - R 3 R 2 R 2 - 6 R 3 - 1 0 2 0 - 9 0 - 2 2 0 - 10 0 0 0 1 3 R 1 R 1 + R 2 1 0 - 2 0 9 0 1 - 1 0 5 0 0 0 1 3 R 1 → - R 1 R 2 → - 2 R 2
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soln4 - SOLUTIONS TO ASSIGNMENT#4 Math 152 1 Consider the...

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