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 | 1 | initial version | posted 2014-07-02 10:32:36 -0600  |
How to find the equation of a plane containing two lines?
On the Exam 1 Review sheet, question #13b asks to find an equation of the plane that contains the two lines p(t)=(3+2t,1=t,4+t) and q(t)=(-2+3t,2t,2+t).
I found the intersection of these lines to be at (1,2,3). Would the next step be to find the cross product of the direction vectors and use the intersection point as my (a,b,c) in the general equation for a plane?
| | 2 | No.2 Revision |
How to find the equation of a plane containing two lines?
On the Exam 1 Review sheet, question #13b asks to find an equation of the plane that contains the two lines p(t)=(3+2t,1=t,4+t) and q(t)=(-2+3t,2t,2+t).
I found the intersection of these lines to be at (1,2,3). Would the next step be to find the cross product of the direction vectors and use the intersection point as my (a,b,c) in the general equation for a plane?
Comment: I think you're on the right track!
| | 3 | retagged |
How to find the equation of a plane containing two lines?
On the Exam 1 Review sheet, question #13b asks to find an equation of the plane that contains the two lines p(t)=(3+2t,1=t,4+t) and q(t)=(-2+3t,2t,2+t).
I found the intersection of these lines to be at (1,2,3). Would the next step be to find the cross product of the direction vectors and use the intersection point as my (a,b,c) in the general equation for a plane?
Comment: I think you're on the right track!
