Finding Parametric Equations for a LineDate: 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/ |
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