# equation of the plane help?

Find an equation of the plane.

The plane that passes through the line of intersection of the planes

x − z = 1 and y + 3z = 3

and is perpendicular to the plane

x + y − 4z = 4

### 2 Answers

- PopeLv 71 month ago
Any plane (save one) including that line of intersection may be written in this form:

(x - z - 1) + k(y + 3z - 3) = 0

x + ky + (3k - 1)z + (-3k - 1) = 0

The dot product of normal vectors of the perpendicular planes must be zero.

<1, k, 3k - 1> · <1, 1, -4> = 0

1 + k - 12k + 4 = 0

k = 5/11

x + ky + (3k - 1)z + (-3k - 1) = 0

x + (5/11)y + [3(5/11) - 1]z + [-3(5/11) - 1] = 0

11x + 5y + 4z - 26 = 0

- az_lenderLv 71 month ago
Let me first find a couple of points on that line...

Let x = 1, z = 0, then y = 3. So A(1,3,0) is on the line.

Let y = 0, z = 1, then x = 2. So B(2,0,1) is on the line.

The given plane x + y - 4z = 4

has as its normal vector <1,1,-4>.

If we start at (1,3,0) and go in the direction of the normal to the given plane, we find C(2,4,-4).

Note that vector AB = <1,-3,1>

and vector BC = <0,4,-5>.

The cross-product of AB and BC will be normal to the plane we are SEEKING. That cross-product is

<11,5,4>.

The equation of the desired plane is

11x + 5y + 4z = D,

where D must be 11*1 + 5*3 + 4*0 = 26.

The equation is 11x + 5y + 4z = 26.