Question

Prove that the two lines x+1/3=y+3/5=z+5/7 and x-2/1=y-4/4=z-6/7 are coplanar?

Answer 1

We showed the two lines meet at `(8/7,4/7,0)` so they're necessarily in the same plane.

Explanation:

I'll guess the slash is the fraction bar:

`(x+1)/3=(y+3)/5=(z+5)/7`

`(x-2)/1=(y-4)/4=(z-6)/7`

We'll assume without proof if a plane includes two points then it includes the line between the two points.

So if we find three points, two of them on the first line, two of them on the second, then both lines are in the same plane. So if the lines meet they're in the same plane. (They can also be parallel and in the same plane, but let's check if they meet first.)

We have four equations in three unknowns, so these may or may not meet in general.

`4x-8=y-4` or `y=4x-4`

`(x+1)/3=(y+3)/5 = (4x-1)/5`

`5x+5 = 12x-3`

`8 = 7x`

`x = 8/7`

`y = 4x-4 = 32/7-28/7=4/7`

We need to solve for `z` using one equation and then check the other to prove they meet.

`7(x+1)=3(z+5) = 3z+15`

`z = 1/3 ( 7x - 8) = 1/3(8-8) = 0`

Checking

`(x-2)/1 = (z-6)/7`

`z = 6 + 7x - 14 =7x-8 = 8-8=0 quad sqrt`

We showed the two lines meet at `(8/7,4/7,0)` so they're necessarily in the same plane.