# planes intersection line

There are 4 points: $A(0, -1, 5)$, $B(1, 3, 3)$, $C(5, 4, 0)$ and $D(3, 0, 4)$. The first plane($\pi_1$) contains the points $A$, $B$ and $C$. The second plane ($\pi_2$) contains the points $A$, $B$ and $D$.

Find the Cartesian equation of the line where $\pi_2$ intersects $\pi_1$.

I just need the concept of how to find that line. Should that be a comparison between $\pi_1$ and $\pi_2$ cartesian equations?

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If you can construct the equation for a plane determined by three points, the equations for those two planes taken together comprises your line. – J. M. May 2 '11 at 18:46
I get $\pi_1$ in Cartesian equation as $2x+y+3z=14$ and $\pi_2$ as $6x-7y-11z=-48$ – Paul S May 2 '11 at 18:54
@Paul S: First equation is correct, but the second is not: if you plug in the coordinates of $D$ into the second equation (which is supposed to represent a plane that contains $D$) you get $6(3) -7(0) - 11(4) = -48$; but that says $18 - 44 = -48$, which is false. – Arturo Magidin May 2 '11 at 18:59
in addition how can I get that in $(x-a_1)/b_1=(y-a_2)/b_2=(z-a_3)/b_3$, where a is the position and b is the direction vector – Paul S May 2 '11 at 19:00
@Paul: Okay, that one is correct. Now that you have both equations, you are trying to find the line that lies in both. Either solve the system of equations $-2x-5y-11z=-50$, $2x+y+3z=14$;, or proceed like Isaac suggests: the line must be orthogonal to both the vector $(2,1,3)$ and the vector $(-2,-5,-11)$; this gives you a direction. Now simply find a point, and you'll have a point and a direction, giving you a line. – Arturo Magidin May 2 '11 at 19:11

Given the two equations of the lines, as you said (in a comment) that you have, in the form $Ex+Fy+Gz=H$ (to borrow Arturo's notation), you know that the vector $\langle E,F,G\rangle$ is orthogonal to the plane. You have one such vector for each plane. Since the line of intersection is in both planes, it is orthogonal to both of these vectors. That means that a vector that is orthogonal to both of the orthogonal-to-the-plane vectors is along the line. The cross-product of the two orthogonal-to-the-plane vectors is orthogonal to both. From this and finding one point that is on the line, you can write a parametric/vector equation of the line of intersection.
Each plane will have an equation of the form $$Ex + Fy + Gz = H$$ for some constants $E$, $F$, $G$, and $H$; a point will lie in the intersection of both planes if and only if it satisfies both equations at the same time, so you'll want to find all solutions to a system of two equations in three unknowns. The set of solutions will give you the line.
To find the equation for a plane given three points that are not collinear, $P$, $Q$, and $R$, you first find the normal vector to the plane. This is a vector that is perpendicular to any vector in the plane. Simplest way to find it is to find two non-collinear vectors that lie in the plane, for example, the vectors $\mathbf{v}_1 = P-Q$ and $\mathbf{v}_2 = R-Q$. With those two vectors, you can compute the normal vector of the plane with $\mathbf{n}=\mathbf{v}_1\times \mathbf{v}_2$ (cross product). If $\mathbf{n}=(a,b,c)$, then the equation of the plane is of the form $ax+by+cz = d$ for some constant $d$. Use the points that are in the plane to find the value of $d$.