Asked by greg
Hi, I'm having trouble answering this question.
The lines l1 and l2 are defined as
l1: (x-1)/3 = (y-5)/2 = (z-12)/-2
l2: (x-1)/8 = (y-5)/11 = (z-12)/6
The plane j contains both l1 and l2. Find the Cartesian equation of j.
The lines l1 and l2 are defined as
l1: (x-1)/3 = (y-5)/2 = (z-12)/-2
l2: (x-1)/8 = (y-5)/11 = (z-12)/6
The plane j contains both l1 and l2. Find the Cartesian equation of j.
Answers
Answered by
Steve
Take the cross product and the normal to both lines is <2,-2,1>
The plane is then just
2(x-1)-2(y-5)+1(z-12) = 0
2x-2y+z = 4
The plane is then just
2(x-1)-2(y-5)+1(z-12) = 0
2x-2y+z = 4
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