Question

1. Give a geometrical interpretation of the intersection of the planes with equations
   x + y − 3 = 0
   y + z + 5 = 0
   x + z + 2 = 0

Answer 1

x+y-3=0, or x+y=3 is a plane A parallel to the z-axis, n₁=<1,1,0>

y+z+5=0, or y+z=-5 is a plane B parallel to the x-axis, n₂=<0,1,1>

x+z+2=0, or x+z=-2 is a plane C parallel to the y-axis, n₃=<1,0,1>, where n represents the normal vector.

We can find the angle θ between the planes by looking at the angle between their normals. For two vectors, a and b, |a||b|cosθ=a·b. |n₁|=|n₂|=|n₃|=√2.

n₁·n₂=n₁·n₃=n₂·n₃=1, therefore 2cosθ=1, θ=π/3 or 60° in each case, so the planes intersect at 60°.

Let x=t, then y=3-t, z=-5-y=-8+t, the parameterised equation of the line intersection of planes A and B;

x=t, y=3-t, z=-2-t, the parameterised equation of the intersection of A and C;

x=t, z=-2-t, y=-5-z=-3+t, the parameterised equation of the intersection of B and C.