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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.
 Spring '08
 Beck
 Equations

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