Question

how to find a set of Parametric Equations for the line of intersection of two non parallel planes

Answer 1

First, we need to represent the equations of the planes; ax+by+cz=d, and ex+fy+gz=h, where a, b, c, d, e, f, g, h are coefficients (constants).

Now we need the normal vectors n1 and n2 for these planes: n1=(a,b,c) and n2=(e,f,g).

The vector product n1 X n2=(bg-cf,ce-ag,af-be).

This is the vector of the line of intersection.

We can find a point in common on both planes, which means that point will be on the line of intersection. All we need to do is put x, y or z to zero and solve for the remaining two variables. Let's put x=0, leaving by+cz=d and fy+gz=h.

fby+fcz=fd 

bfy+bgz=bh

Subtract: and solve for z=(fd-bh)/(fc-bg).

Now we can find y: y=(d-cz)/b=(d-c(fd-bh)/(fc-bg))/b=(dfc-dbg-cfd+cbh)/(b(fc-bg))=(ch-dg)/(fc-bg).

The point on the line is therefore (x,y,z)=(0,(ch-dg)/(fc-bg),(fd-bh)/(fc-bg)). Call this point A.

We can now use this point and n1 X n2 with parameter t to write the parametrised equations:

 W(t)=A+(n1 X n2)t=(0,(ch-dg)/(fc-bg),(fd-bh)/(fc-bg))+(bg-cf,ce-ag,af-be)t, from which:

x=(bg-cf)t, y=(ch-dg)/(fc-bg)+(ce-ag)t, z=(fd-bh)/(fc-bg)+(af-be)t.

 

Answer 2

Find parametric equations for the line of intersection of the planes x + y − z = 1 and 3x + 2y − z = 0.

Also find the angle between these two planes.

To find a point on this line we can for instance set z = 0 and then use the above equations to solve for x
and y. In this case we get x = −2 and y = 3 so (−2, 3, 0) is a point on the line. Also the direction of the line
lives in both planes and so in particular is perpendicular to both normal vectors, therefore a vector which is
parallel to the line is given by (1, 1, −1) × (3, 2, −1) = (1, −2, −1).
 

Thus an equation of the line is given by the vector equation

(x, y, z) = (−2, 3, 0) + t(1, −2, −1),

Or the parametric equations

x = −2 + t, y = 3 − 2t, z = −t,

Or the symmetric equations

x + 2 =y − 3−2 = −z.

Then angle between these planes is given by cos θ = |(1, 1, −1)|^(-1) |(3, 2, −1)|^(-1) (1, 1, −1) · (3, 2, −1) =6/√42, and so θ = cos−1(6/√42).

Also understand about Parametric Equation of a Circle