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 June 9th, 2014, 07:16 AM #1 Newbie   Joined: Jun 2014 From: Iowa Posts: 5 Thanks: 0 Planes and lines in 3D Hi everyone ! I'd like to know if someone could help me : If we take two planes $P_{1}\, : \, ax+by+cz+d=0$ and $P_{2}\, : \, a'x+b'y+c'z+d'=0$, and call $D=P_{1} \cap P_{2}$ (D is a line). How can we prove that a random plan P contains D if and only if P has the following cartesian equation : $\lambda (a'x+b'y+c'z+d') + \mu (ax+by+cz+d)=0$ Thanks a lot !!
 June 9th, 2014, 08:50 AM #2 Newbie   Joined: Jun 2014 From: Colombia Posts: 4 Thanks: 0 Hi, it is clear that any point whose satisfy both equations $P_{1}\, : \, ax+by+cz+d=0$ and $P_{2}\, : \, a'x+b'y+c'z+d'=0$ lies on their line of intersection. Clearly the coordinates of such point satisfy the equation $\lambda (a'x+b'y+c'z+d') + \mu (ax+by+cz+d)=0$, where $\lambda$ and $\mu$ are arbitrary constants which may assume all real values except that of them may not be zero simultaneously. That equation represents a family of planes passing through the intersection of given planes.
 June 11th, 2014, 03:19 AM #3 Newbie   Joined: Jun 2014 From: Iowa Posts: 5 Thanks: 0 Thanks ! And how do you find the equation of a plan P which contains a line D and a point M(a,b,c) (Assuming M is not on the line D)?
 June 11th, 2014, 09:17 PM #4 Newbie   Joined: Jun 2014 From: Iowa Posts: 5 Thanks: 0 Anyone help? thanks a lot
 June 13th, 2014, 02:00 AM #5 Banned Camp   Joined: May 2014 From: USA Posts: 3 Thanks: 0 Take help from online tutoring .There are various sites which are very cheap try it .
 June 13th, 2014, 03:03 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,099 Thanks: 703 Math Focus: Physics, mathematical modelling, numerical and computational solutions Ignore the previous post; there is nothing wrong with you asking the question and we'll happily help. I'll see if I can get an answer to you soon (unless someone else on the forum beats me to it )
 June 13th, 2014, 11:04 AM #7 Newbie   Joined: Jun 2014 From: Colombia Posts: 4 Thanks: 0 ok, just wait. you have the line and the point then you have three non-collinear points $P_1, P_2, P_3$ then you have three equations and 4 variables, then you solve the system for any three of the variables in terms of the fourth (which is not be zero). substituting these values in the general equation ($Ax+By+Cz+D=0$) next dividing through the letter and you have the equation.

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