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theorist
Apr27-04, 01:44 AM
Hello,

How do I find the line of intersections of the two planes 7x - 2y + 3z = -2 and -3x + y + 2z + 5 =0, without having to resort to solving it by row reduction?

cookiemonster
Apr27-04, 02:09 AM
Calculate the cross product of the two normal vectors of the planes. Then plug in one point of intersection to yield the line of intersection.

cookiemonster

HallsofIvy
Apr27-04, 06:58 AM
Or, if that is too much trouble, solve the two equations for two of the variables, leaving the third as parameter: from 7x - 2y + 3z = -2 and -3x + y + 2z + 5 =0, multiply the second equation by 2 and add to get
(7+2(-3))x+ (-2+2)y+ (3+2(2))z+ 2(5)= -2 or x+ 7z+ 10= -2. From that,
x= -7z- 12 and then, using the second equation, -3(-7z-12)+ y+ 2z+ 5= 0 so
y= -23z- 41 or, writing the parameter as "t"
x= -7t- 12, y= -23t- 41, z= t is the line of intersection.