Spring 2007 - Linshaw's Class - Quiz 1

# Spring 2007 - Linshaw's Class - Quiz 1 - v 1 = a 2 3 − 1...

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Name: PID: TA: Sec. No: Sec. Time: Math 20C. Quiz 1 April 13, 2007 1. Find an equation for the plane which passes through the points P = (3 , 1 , 2) , Q = (8 , 2 , 4) , R = (0 , 2 , 3) . Solution: The vectors −→ PQ = a 5 , 3 , 2 A and −→ PR = a− 3 , 1 , 1 A lie in the plane, so a normal vector n is given by −→ PQ × −→ PR = a 5 , 11 , 4 A . The plane contains P and is normal to n , so the equation for the plane is 5( x 3) 11( y + 1) + 4( z 2) = 0, or equivalently, 5 x 11 y + 4 z 34 = 0. 2. Consider the lines L 1 and L 2 given parametrically as follows: L 1 : x = 1 + 2 t, y = 3 t, z = 2 t, L 2 : x = 1 + s, y = 4 + s, z = 1 + 3 s. Determine if L 1 and L 2 are parallel, skew, or intersecting. If they intersect, ±nd the point of intersection. Solution: L 1 is parallel to the vector
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Unformatted text preview: v 1 = a 2 , 3 , − 1 A , and L 2 is parallel to the vector v 2 = a 1 , 1 , 3 A . Since these vectors are not parallel, it follows that L 1 and L 2 are not parallel. To see whether they intersect, we need to ±nd a solution (if any) to the linear system: 1 + 2 t = − 1 + s, 3 t = 4 + s, 2 − t = 1 + 3 s. Solving the ±rst two equations for t and s yields t = 6 and s = 14. But the third equation is not satis±ed, so we conclude that this system has no solution. Therefore the lines are skew (ie, they do not intersect)....
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## This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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