Plane question?

Question

Find the equation of the plane containing the line L_1, and parallel to the line L_2, where:

L_1: [x,y,z]^T = [ -1,1,-3 ]^T + t [ 1,-3,0 ]^T.

L_2 is the intersection of the planes

2 x +3 y +2 z = -4 and 2 x - 1 y +1 z = 5.

Should be in the form Ax + By + Cz = D. Adjust your coefficients so that A, B, C and D are in lowest terms (have 1 as the greatest common factor), and D is positive.

A =

B =

C =

D =

Answer 1

I propose doing this the easy way.

Above all, don't actually compute L2. Just find the direction vector. We can do this by taking the normal vectors of the two planes (which we can write down from the equations: (2, 3, 4) and (2, -1, 1)) and finding their cross product. This vector is normal to both, so is parallel to both planes and hence to their intersection.

So L2 is parallel to (2, 3, 4)×(2, -1, 1) = (7, 6, -8).
Now the desired plane is parallel to the direction vectors of L1 and L2, so their cross product is normal to both of these and hence is the normal vector of the plane. So we get
n = (1, -3, 0)×(7, 6, -8) = (24, 8, 27).

So the equation is 24x + 8y + 27z = D for some D. Taking t=0 we know (-1, 1, -3) is in the plane, so D = -24 + 8 - 81 = -97. So the equation of the plane is
24x + 8y + 27z = -97.

Answer 2

First let's find the directional vector v of line L_2. This can be found by taking the cross product of the normal vectors of the two given planes.

<2, 3, 2> X <2, -1, 1> = <5, 2, -8>

The plane we want contains the directional vector from line L_1 and the just found directional vector of line L_2. This along with the given point (-1,1,-3) is enough to define the plane.

The equation of the desired plane in vector form is:

P = <-1, 1, -3> + t<1, -3, 0> + s<5, 2, -8>
where s and t are scalars ranging over the real numbers

Now let's convert its form to the desired form. Take the cross product of the two vectors lieing in the plane to obtain the normal vector n, to the plane.

n = <1, -3, 0> X <5, 2, -8> = <24, 8, 17>

One of the points in the plane is (-1, 1, -3). This along with the normal vector is sufficient to write the equation of the plane.

24(x + 1) + 8(y - 1) + 17(z + 3) = 0
24x + 24 + 8y - 8 + 17z + 51 = 0
24x + 8y + 17z = -67

Multiply thru by -1 so D is positive.

- 24x - 8y - 17z = 67

A = -24
B = -8
C = -17
D = 67

Answer 3

> L_2 is the intersection of the planes
> 2 x +3 y +2 z = -4 and 2 x - 1 y +1 z = 5.

[ 2,3,2 ]^T is then perpendicular to the first plane and [ 2,-1,1 ]^T is perpendicular to the second plane, so their cross product gives the direction of the line of intersection: [ 5,2,-8 ]^T.

Then, since L_1 is in the unknown plane, the cross product of [ 1,-3,0 ]^T and [ 5,2,-8 ]^T is perpendicular to the plane. That second cross product is [ 24, 8, 17 ]^T. Since [ -1,1,-3 ]^T is a point in the plane you have:

24 (x - -1) + 8 (y - 1) + 17 (z - -3) = 0
24x + 24 + 8y - 8 + 17z + 51 = 0
24x + 8y + 17z + 67 = 0
24x + 8y + 17z = -67
-24x - 8y - 17z = 67

A = -24
B = -8
C = -17
D = 67

Dan