Find the condition for the two lines to be coplanar.
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!
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/
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.
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
I guess now I understand how to solve it.
Thanks for the resource!
To prove that two lines are coplanar, we need to show that they lie on the same plane. In other words, we need to show that there is at least one common point between the two lines.
Let's consider the two given lines:
Line 1: a1 + b1x + c1x + d1 = 0
Line 2: a2 + b2x + c2x + d2 = 0
To find the condition for these two lines to be coplanar, we need to find a common point that satisfies both equations.
We can rewrite Line 1 and Line 2 as follows:
Line 1: x = (-(b1+d1))/(a1+c1)
Line 2: x = (-(b2+d2))/(a2+c2)
As both x-values are equal, we have:
(-(b1+d1))/(a1+c1) = (-(b2+d2))/(a2+c2)
Cross-multiplying and simplifying, we get:
(a2+c2) * (-(b1+d1)) = (a1+c1) * (-(b2+d2))
Expanding the above expression, we have:
a2*(-b1) + a2*(-d1) + c2*(-b1) + c2*(-d1) = a1*(-b2) + a1*(-d2) + c1*(-b2) + c1*(-d2)
Simplifying further, we get:
(-a2*b1) + (-a2*d1) + (-c2*b1) + (-c2*d1) = (-a1*b2) + (-a1*d2) + (-c1*b2) + (-c1*d2)
Now, we can group the terms and rearrange them:
(-a2*b1 - c2*b1) + (-a2*d1 - c2*d1) = (-a1*b2 - c1*b2) + (-a1*d2 - c1*d2)
Combining like terms, we get:
(-a2 - c2) * b1 + (-a2 - c2) * d1 = (-a1 - c1) * b2 + (-a1 - c1) * d2
If we define new coefficients as follows:
A = -a2 - c2
B = -a1 - c1
C = -b1
D = -b2
E = -d1
F = -d2
Then the equation becomes:
A * C + A * E = B * D + B * F
Simplifying further:
A*(C + E) = B*(D + F)
Now, to ensure that there exists a common point that satisfies both equations, we need A and B to be non-zero. This can be expressed as:
A ≠ 0 and B ≠ 0
Taking A and B back to their original form:
-a2 - c2 ≠ 0 and -a1 - c1 ≠ 0
Finally, the condition for the two lines to be coplanar is:
a2 + c2 ≠ 0 and a1 + c1 ≠ 0