Find a plane through P (2,1,1) and perpendicular to the line of intersection of the planes: 2x+y-z = 3 and x+2y+z = 2.

Answer 1

r = r₀ + vt is the equation for the line of intersection.

We acquire v from the cross-product between normal vectors on the planes, which will be in the direction of the intersection line.

2x + y -z = 3, n₁ = < 2,1,-1 >

x + 2y + z = 2, n₂ = < 1, 2, 1>

I. v = n₁ x n₂ = 3i - 3j + 3k = < 3, -3, 3>

To find a point on the line, set z = 0 in the system of equations and solve

2x + y = 3

x + 2y = 2

x = 2 - 2y

2(2-2y) + y = 3

4 - 4y + y = 3

4 - 3 = 3y

y = 1/3

x = 4/3

II. r₀ = <4/3, 1/3, 0>

Using I and II and substituting into the original equation

r = < 4/3,1/3,0 > + < 3,-3,3 >t

Where we have the original point, and when t = 1 for example, it is directed v units to the other point on the line intersecting each plane.

r(t=1) = coefficients normal to the plane = (13/3, -8/3, 3)

13x - 8y + 9z = 3d

3d = 27

13x - 8y + 9z = 27