Help with Equation of planes containing lines???

Question

3) Find an equation of the plane that contains the line:

x=2 - 7t, y=3 - t, z=8 plus 4t

and is parallel to the plane -7x plus 53y plus z=7

Leave your answer in the form Ax plus By plus Cz = D, where A, B, C are integers.

4) Find an equation of the plane that passes through the point (6, -6, 7) and contains the line: x=5 - 7t, y=4 plus t, z=1 plus 6t

Answer 1

1.

The components of the normal to the given plane are the coefficients of the eq. of the plane, so it's n=< -7, 53, 1>. We want a plane that is parallel to this one, so the normal of the new plane must be parallel to the normal of the original plane. So we can pick n itself as the normal of the new plane. Thus the eq of the new plane is of the form:

-7x +53y +z = d.  To determine d, plug in one point that is on the line and thus also in the plane we want. When t=0, the line passes through the point (2,3,8). Substituting gives d = 153.

So the answer is -7x+53y+z=153.

As a check, replace x, y, and z by the parametric form for the line:

-7(2-7t) +53(3-t) +(8+4t) which = 153. So it's correct. (The check tells us that every point on the line is also in the new plane).

2.We can write the eq of a plane once we have a normal vector and one point in the plane. So we need a normal. The best way to do that is to take the cross product of 2 vectors in the plane we want. The line lies in the plane, and the parametric eq. tells us that the line is parallel to the vector v=<-7,1,6>, so v is one vector in the plane. To get a 2nd vector in the plane, we need 2 points in the plane. We are given one point P(6,-6,7). To get another point, we note that when t=0, the line passes through Q(5,4,1) and Q must be in the plane we want. Form the vector PQ=<5-6, 4-(-6),1-7> = <-1, 10, -6>. (PQ is of course in the plane we want.) Now take the cross product PQ x v = <66, 48, 69> = n.

Now we have a point P in the plane and a normal to the plane n. The eq of a plane is

a(X-Xo)+b(Y-Yo)+c(Z-Zo)=0 where  the normal to the plane=<a,b,c> and (Xo,Yo,Zo) is any point on the plane.

So the answer is    66(x-6)+48(y+6)+69(z-7)=0.

As a check, if you replace x, y, and z in this equation by the parametric form of the line, you see it works (which means that all points on the line are indeed in this plane).