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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,
Jeff
```

```
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
or
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
http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra

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