MATH 126 Midterm: MATH 126 UW 126 Winter 10 Sol1
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That is they do not intersect and they are not parallel. 1 but then comparing second coordinates 5 = 4a = 4 is wrong. They do not intersect because if we try to solve. < 2 t, 3 + 5t, 6t >=< 3 + s, 1 + 4s, 2 + 3s > we get s = 1 t from the rst component. Plugging that into the second we get 3 + 5t = 1 + 4( 1 t) = 3 4t so t = 6/9 and s = 6/9 1 = 15/9 than the last component gives. 36/9 = 63/9 which is false. (b) find the distance between them. First we need a vector normal to both vectors: n =< 1, 5, 6 > < 1, 4, 3 >=< 9, 9, 9 > or it easier to work with the parallel vector < 1, 1, 1 >. Then the distance between the two lines is comp< 1,1, 1> (cid:12)(cid:12)(cid:12)
