Question

show that each of the pairs of lines x-y+z-5=0, x-3y+6=0 and 2y+z-5=0,12x-17y+4z-3=0

Answer 1

The planes x-y+z-5=0 and x-3y+6=0 intersect along a line L1.

The planes 2y+z-5=0 and 12x-17y+4z-3=0 intersect along a line L2.

For L1, let y=t (parameter), then x-3t+6=0, x=3t-6, and 3t-6-t+z-5=0,

z=11-2t. So t=(x+6)/3=y=(11-z)/2 defines L1.

For L2, let z=t, then y=(5-t)/2 and 12x-17(5-t)/2+4t-3=0,

12x-85/2+17t/2+4t-3=0, 

24x=91-25t, x=(91-25t)/24.

So t=(91-24x)/25=5-2y=z defines L2.

It follows that the position vectors of a point P on L1 are <3t-6,t,11-2t> and a point Q on L2 has position vector <(91-25t)/24,(5-t)/2,t>

Arbitrarily, when t=0 we get position vectors P=<-6,0,11> on L1 and Q=<91/24,5/2,0>; when t=1 we get:

R=<-3,1,9> on L1 and S=<11/4,2,1> on L2.

L1=<-6,0,11>+λ₁<3,1-2> and L2=<91/24,5/2,0>+λ₂<-25/24,-½,1> is another way of defining the lines, where the λ's are scalars (any number).

The question seems incomplete: "show that..." so it seems logical to check whether the lines intersect. We use the line equations to do this.

If the lines intersect, there will be one set of coordinates (x,y,z) which satisfy both line equations.

-6+3λ₁=91/24-25λ₂/24

λ₁=5/2-λ₂/2

11-2λ₁=λ₂

In this system of 3 equations there are only 2 variables: λ₁ and λ₂. Substituting for λ₂ in the second equation we get:

λ₁=5/2-11/2+λ₁, 0=-3 which is false, so there is no solution to the system, so the lines don't intersect.

We can use the direction vectors (derived from the parametric equations) of the lines to determine if they are parallel:

For L1: x=-6+3t, y=t, z=11-2t; direction vector, v₁=<3,1,-2>.

For L2: x=91/24-25t/24, y=5/2-½t, z=t; direction vector, v₂=<-25/24,-½,1>.

Since v₂≠μv₁ for some scalar value, μ the lines are not parallel. v₁.v₂=-25/8-1/2-2≠0, so neither are they perpendicular.