Find the condition for the two lines to be coplanar.

Question

a1 + b1x + c1x + d1 =0 = a2 + b2x + c2x + d2
a3 + b3x + c3x + d3 =0 = a4 + b4x + c4x + d4

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I know that the general line passing through the first lines can be represented as a1 + b1x + c1x + d1 +k(a2 + b2x + c2x + d2), for some real number k

Also, the general line passing through the second lines can be represented as a3 + b3x + c3x + d3 +d(a4 + b4x + c4x + d4 ), for some real number r

How do we then prove the required result?

Thank you!

Answer 1

I don't quite get your notation, but there are lots of discussions of this topic online. You might want to start here:

byjus.com/maths/coplanarity-two-lines/

Answer 2

There's a typo in the question.

The question should be like this:

Find the condition for the two lines to be coplanar.
a1x + b1y + c1z + d1 =0 = a2x + b2y + c2z + d2

a3x + b3y + c3z + d3 =0 = a4x + b4y + c4z + d4

Answer 3

I guess now I understand how to solve it.
Thanks for the resource!

Answer 4

Also befuddled by your notation, but the simplest method to see
if 2 lines are coplanar can be illustrated with this example:

suppose we have
L1 = (2,3,1) + s(4,-1,2) and
L2 = (1,0,2) + t(1,2,3)

let's find a plane perpendicular to both lines,
this is done by finding the cross product of <4,-1,2> and <1,2,3>
which would be <7,10,-9> in simplest form, and would be normal to the plane.

So the plane would look like this:
7x + 10y - 9z = k, where k is a constant
let L2 lie on this plane, then using the point(1,0,2)
7(1) + 10(0) - 9(2) = k
k = -11

so the plane containing L2 is 7x + 10y - 9z = -11
we know L1 is perpendicular to our plane, but is (2,3,1) on it ??
Is 7(2) + 10(3) - 9(1) = -11, NO

so the 2 lines are NOT coplanar.