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I have no idea what to do for this problem.

Find an equation for the plane containing the line L:(x-4)/(-4) =(z+3)/2; y = -3 and the point Q(-3,2,-2).

The equation, y= -3 is getting me confused. I have no idea what to with this equation or how to complete the problem with it. Any help would be greatly appreciated.

First I'll answer your y=-3 question and then derive the plane eqn. The eqn for a line in 3-D is given by the so-called symmetric eqns which come from the 3 eqns for the components of the line along the 3 axes:

x = x0 + at

y = y0 + bt

z = z0 + ct

where (x0,y0,z0) is a point in 3-D and <a,b,c> are the so-called directon numbers. These are just amplitudes along the axes proportional to a free parameter t (increase or decrease t to obtain any point on the line; eg., t = 0 -> x0,y0,z0). If these eqns are each solved for t, you get (for the x component)

(x-x0)/a = t.

Note however, that if any of the a,b,c are 0, then you can't divided by it. In the present case, b = 0 so that you are left with y = y0 = -3, which is fine.

Deriving the eqn for the plane is a good exercize. First, carefully note the wording of the question: the plane contains the given line and also contains the given point (but the line doesn't necessarily contain the point; in fact it doesn't, you should confirm this). Thus we want to find a plane that is a) parallel to the line and b) contains the given point, Q (plane only). The key here is to use the handy definition of a plane as all the points perpindicular to a line, N; by definition of the dot product

N･V = 0 = N･<v1,v2,v3> for some vector v in the plane.

What we will do is find the vector perpindicular to 2 vectors we know are in the plane and use the direction numers of this normal vector to write out the so-called scalar eqn of the plane.

The first vector comes from the symmetric eqns of the line, L, which by definition are <-4,0,2>. The 2nd vector is obtained by noting that the point in the eqns for the line is P = (4,0,-3) and the given point, Q = (-3,2,-2), are both points in the plane; thus the vector between them is as well

PQ = (px-qx, py-qy, pz-qz) = (4+3,0-2,-3+2) = (7,-2,-1).

The vector, N, perpindicular to PQ, is given by the cross product of the 2 in-plane vectors (you should know this)

N = L x PQ = 4i + 10j +8k, where i, j, and k are the unit vectors for x,y,z (you should know how to take a cross product). From this, the components of N are Nx = 4, Ny = 10 and Nz = 8. By definition, the scalar eqn for the plane is

N･(X-Q) (X = (x,y,z))

or

4x + 10y + 8z = -N･Q = -{4･(-3) + 10･(2) + 8･(-2)) = 8.

For exercise, you should confirm that N･L = 0 and N･PQ = 0 give consistent equations for the direction numbers <nx,ny,nz>.

Randy

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