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Finding Parametric Equations for a Line

Date: 05/24/2000 at 00:27:17
From: Jeffrey
Subject: Vectors - parametric equations

Find the angle between the two planes given by:

     x - 2y + z = 0   and   2x + 3y - 2z = 0

and find the parametric equations for their line of intersection.

I have tried to solve this a number of times only to find myself 
stuck. I am studying first year engineering and I need to solve this 
problem to pass my first semester.

Yours sincerely,

Date: 05/24/2000 at 13:03:19
From: Doctor Rob
Subject: Re: Vectors - parametric equations

Thanks for writing to Ask Dr. Math, Jeffrey.

The normal vectors to the two planes are:

     v1 = (1,-2,1)   and   v2 = (2,3,-2)

(These are found from the coefficients of x, y, and z, respectively, 
of the equations of the planes.) The angle between the planes is the 
same as the angle between their normals, which can be found from:

     cos(theta) = v1.v2/(|v1|*|v2|)

(Actually the planes form two angles, one of which is the above and 
the other is 180 degrees minus the above angle.)

The direction of their line of intersection is along the vector:

     v3 = v1 x v2

That is because the line of intersection lies in each plane, and so is 
perpendicular to each normal vector, and their cross-product also is 
perpendicular to both the normal vectors. You can compute the cross 
product v3.

If v3 = (a,b,c), and v0 = (x0,y0,z0) is a point on the line, then the 
parametric form of the equation of the line is:

     v = v0 + v3*t
     x = x0 + a*t
     y = y0 + b*t
     z = z0 + c*t

where t is the parameter, and ranges over the whole real line. Now use 
the fact that the point (0,0,0) lies in both planes, and so on the 
line of intersection, so v0 = (0,0,0) is a valid choice.

- Doctor Rob, The Math Forum   
Associated Topics:
College Linear Algebra

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