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Notes001(22) - We know that the plane will have an equation of the form a x& 5 b y& 0 c z& 0 = 0 or ax by cz = 5 a Since the point Y(0&

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September 2, 2009 NAME____________________________________________ Find an equation for the plane that has x intercept X (5 ; 0 ; 0) , y intercept Y (0 ; 3 ; 0) , and z intercept Z (0 ; 0 ; 8) . Solution 1 (using the cross product): The vectors XY = 5 ; 3 ; 0 i and XZ = 5 ; 0 ; 8 i are both parallel to the plane in question. Thus a normal vector to this plane is n = XY ± XZ = i j k 5 3 0 5 0 8 = 24 i + 40 j 15 k . An equation for the plane is thus 24 ( x 5) + 40 ( y 0) 15 ( z 0) = 0 which we can also write as 24 ( x 5) 40 ( y 0) + 15 ( z 0) = 0 or as 24 x 40 y + 15 z = 120 . Solution 2 (not using the cross product):
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Unformatted text preview: We know that the plane will have an equation of the form a ( x & 5) + b ( y & 0) + c ( z & 0) = 0 or ax + by + cz = 5 a . Since the point Y (0 ; & 3 ; 0) must be in this plane, we must have & 3 b = 5 a . Since the point Z (0 ; ; 8) must be in this plane, we must have 8 c = 5 a . Thus b = & 5 3 a and c = 5 8 a . If we choose a = 24 , we obtain b = & 40 and c = 15 : We thus obtain the same solution as given in Solution 1. 1...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.

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