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**Unformatted text preview: **HOMEWORK 2 SOLUTIONS Levandosky, Linear Algebra 2.11 There are many possible parameterizations. Following the notation in the text, we can first let x = 3 1 , and obtain a direction vector using the vector whose tail is at (2 ,- 3) and head is at (3 , 1): v = 3- 2 1- (- 3) = 1 4 . Thus, the line can be written parametrically as L = 3 1 + t 1 4 | t R . 2.13 As above, one such parameterization is given by x = 1 2 3 and v = 1- (- 2) 2- 1 3- 2 = 3 1 1 ; i.e. L = 1 2 3 + t 3 1 1 | t R . 2.15 We can first let x = 1 2 3 . To obtain two direction vectors, we can use the vectors whose tails are at (1 , 2 , 3) and whose heads are at (2 , 3 , 4) and (2 , 1 , 5). This gives v 1 = 2- 1 3- 2 4- 3 = 1 1 1 and v 2 = 2- 1 1- 2 5- 3 = 1- 1 2 . Thus, the plane can be written paramterically as P = 1 2 3 + s 1 1 1 + t 1- 1 2 | s, t R . 2.17 We can first let x = 2 3 4 5 . To obtain three direction vectors, we can use the vectors whose tails are at (2 , 3 , 4 , 5) and whose heads are at (1 , 1 , 1 , 1), 1 2 HOMEWORK 2 SOLUTIONS (0 , 1 , , 1), and (- 1 ,- 2 , 3 , 1). This gives v 1 = 1- 2 1- 3 1- 4 1- 5 = - 1- 2- 3- 4 , v 2 = - 2 1- 3- 4 1- 5 = - 2- 2- 4- 4 , and v 3 = - 1- 2- 2- 3 3- 4 1- 5 = - 3- 5- 1- 4 . Thus, the hyperplane can be written paramterically as P = 2 3 4 5 + r - 1- 2- 3- 4 + s - 2- 2- 4- 4 + t - 3- 5- 1- 4 | r, s, t R ....

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