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Unformatted text preview: Solutions to Prelim 2, Math 192, Fall 2002 1. (a) The plane should contain the points (1 , , 1) and (0 ,- 1 , 0) and be parallel to the normal vector of x- y + z = 0, i.e. i- j + k . So, a normal vector is- 2 i + 2 k , the cross product of i- j + k and i + j + k , the vector between the two given points. So, an equation for the plane is- x + z = 0. (b) The given lines are parallel to the vectors 2 j- k and 2 i +3 k , whose cross product is 6 i- 2 j- 4 k , a vector perpendicular to both lines. (c) The plane must contain the points (1 ,- 2 , 0) and (2 , , 2) and be parallel to the vectors- i + 3 j- 2 k and 3 i- 9 j + 6 k . Notice that these vectors are parallel. As in part (a), we take the cross product of- i +3 j- 2 k and i +2 j +2 k , the vector between the points, to get the normal vector n = 10 i- 5 k . Then an equation for the plane is 10 x- 5 z = 10. 2. (a) We compute v ( t ) =- e t j + 3 cos( 3 t ) k and a ( t ) =- e t j- 3 sin( 3 t ) k ....
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- Spring '06