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Unformatted text preview: x , y , and z . Namely, we get the vector h 1 , 1 , 1 i . Furthermore, we know that since our line is perpendicular to the line x = 1 + t , y = 1t , z = 2 t , it is also perpendicular to the direction vector of that line. To ﬁnd the direction vector, we simply look at the coeﬃcients in front of the t ’s. Thus, the direction vector for this line is h 1 ,1 , 2 i . We now have two vectors that our line (or, more speciﬁcally, our line’s direction vector) is perpendicular to. Thus, we may simply compute the cross product h 1 , 1 , 1 i × h 1 ,1 , 2 i to ﬁnd a suitable direction vector for our line. ± ± ± ± ± ± i j k 1 1 1 11 2 ± ± ± ± ± ± = 3 i1 j2 k Now let’s plug this into our formula for a parametrized line. When we do, we get x = 3 t, y = 1t, z = 22 t. 1...
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This homework help was uploaded on 04/17/2008 for the course MATH 231 taught by Professor Bociu during the Spring '08 term at UVA.
 Spring '08
 BOCIU
 Calculus, Vectors

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