Find an equation of a plane that contains the origin & is perpendicular to each of the planes.?

Question

Fin an equation of a plane that contains the origin and is perpendicular to each of the planes x+y+z=4 and x-y+z=4. Find the pt of intersection of the three planes.

for the first answer i got 2x - 2z but the answer is x-z=0 so i don't know how to that answer. since i didn't get the first answer i didn't bother trying to get the second answer. please explain how to get the answers.

Answer 1

Let P1 and P2 denote the given planes. Let P3 denote the plane to be found.

P1: x + y + z = 4
P2: x - y + z = 4

P3 is orthogonal to both P1 and P2. Find its normal vector by taking the cross product of normal vectors for P1 and P2.

<1, 1, 1> x <1, -1, 1> = <2, 0, -2>

The equation of the P3 is 2x - 2z = d. The equation passes through the origin or (0, 0, 0). Substitute this point in the equation for P3 to get d = 0. Also, divide through by 2 to give

P3: x - z = 0

To find the point of intersection, solve the linear system of the three planes:

[1, 1, 1] [x] [4]
[1, -1, 1] [y] = [4]
[1, 0, -1] [z] [0]

The solution for point of intersection is [2, 0, 2]^T.

Answer 2

Find an equation of a plane that contains the origin and is perpendicular to each of the planes x + y + z = 4 and
x - y + z = 4. Find the point of intersection of the three planes.

The normal vector n, of the desired plane will be perpendicular to the two normal vectors, n1 and n2, of the given planes.

n1 = <1, 1, 1>
n2 = <1, -1, 1>

Take the cross product of n1 and n2 to get n.

n = n1 X n2 = <1, 1, 1> X <1, -1, 1> = <2, 0, -2>

Any non-zero scalar multiple of n will also be a normal vector to the desired plane. Divide by 2.

n = <1, 0, -1>

With the normal vector n and a point on the plane (0,0,0) we can write the equation of the plane.

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

(Please note 2x - 2z = 0 is also an equation of the plane. If you have a Cartesian equation of a plane, you can multiply all the coefficients by any non-zero scalar and you will still have an equation of the plane.)
______________

Now find the point of intersection of the three planes.

P1: x + y + z = 4
P2: x - y + z = 4
P3: x - z = 0

Add the first two equations together.

P1 + P2: 2x + 2z = 8
x + z = 4

Now add this result to the third equation.

(1/2)(P1 + P2) + P3: 2x = 4
x = 2

Substitute into equation 3.

P3: x - z = 0
2 - z = 0
z = 2

Substitute into equation 1.

P1: x + y + z = 4
2 + y + 2 = 4
y = 0

The point of intersection of the three planes is (2, 0, 2).

Answer 3

commonly used vectors to the intersecting planes are <2,3,a million> and <5,-2,2>. the line of intersection could be perpendicular to the two one in all those normals, so this is parallel to their pass product. |i j ok| |2 3 a million| = <8,a million,-19> |5 -2 2| next, come across a factor on the line undemanding to the intersecting planes. try this by removing between the variables, say, z: 5x - 2y + 2z = -4 4x + 6y + 2z = 4 ---------------------------- x - 8y = -8 a answer to it fairly is x = 0, y = a million; then z = -a million. So the factor (0,a million,-a million) is interior the plane we are in seek of. considering that (0,0,0) is likewise to be in this plane, the vector from (0,0,0) to (0,a million,-a million), that's <0,a million,-a million>, is parallel to the plane. So the pass made from this vector and the vector parallel to the line of intersection is perpendicular to the plane being sought; this pass product is |i j   ok| |8 a million -19| = <18,8,8> |0 a million  -a million| So an equation for the plane is eighteen(x-0) + 8(y-0) + 8(z-0) = 0, or after dividing by 2, 9x + 4y + 4z = 0