quiz2sol

# quiz2sol - equations obtained from the planes. This would...

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Math 251, Spring 2008, Quiz 2 Solutions You are given the following equations: ( plane 1) P 1 : x + y + z = 1 ( plane 2) P 2 : x - y + z = 1 ( line ) L : x = 3 + 2 t ; y = 2 - 3 t ; z = 1 - 4 t Find the equation of a plane that contains the line of intersection of P 1 and P 2 and the point of intersection of P 2 and L . Does your answer look familiar? If yes, justify. Please draw an approximate picture of the problem to get full credit. Solution: For this one there was an easy solution. If you drew a picture, you would have realized that plane P 2 was going through both the line of intersection and the point speciﬁed. Hence P 2 is the right answer! The longer way to do this was: For the line of intersection the cross product of the normal vectors of both the planes would give the direction vector of the line, i.e < 1 , 1 , 1 > × < 1 , - 1 , 1 > = < 2 , 0 , 2 > . A point on this line can be obtained by setting z = 0 and solving the
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Unformatted text preview: equations obtained from the planes. This would yeild the point M = (1 , , 0). The point of intersection of P 2 and L is given by plugging in the expressions for x, y, z from L into the equation of the plane which gives that t =-1 and the required point is N = (1 , 5 , 5). Then the normal vector of the plane you are trying to ﬁnd is given by the cross product of the direction vector of the line and the vector ~ NM which is < 1 ,-1 , 1 > . You also have a point of the plane and so the equation would turn out to be that of P 2. I think almost all gave up after ﬁnding P 1 ∩ P 2 and P 2 ∩ L . This problem was equivalent to two homework problems put together but I think it should have been doable in 25 min. 1...
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## This note was uploaded on 08/10/2008 for the course MATH 251 taught by Professor Beck during the Spring '08 term at Rutgers.

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