Equation of a plane containing two lines?

Question

How would I go about finding the equation of a plane containing the lines:
x-3=4t, y-4=t, z-1=0
and x+1=12t, y-7=6t, z-5=3t

I'm stumped. Thanks.

Answer 1

The two lines can be contained in one plane only if:

i) the lines are parallel, or
ii) the lines intersect

The two lines are:

x - 3 = 4t, y - 4 = t, z - 1 = 0
x + 1 = 12t, y - 7 = 6t, z - 5 = 3t

Let's rewrite the equations in vector form. It's better not to use the variable t, in both equations because they are not the same t.

L1(s) = <3, 4, 1> + s<4, 1, 0>
L2(t) = <-1, 7, 5> + t<12, 6, 3>

The directional vectors of the lines are not parallel so the lines are not parallel. Therefore we need to see if the lines intersect. Solve for s and t to find the point of intersection, if any.

x = 3 + 4s = -1 + 12t
y = 4 + s = 7 + 6t
z = 1 = 5 + 3t

z: 1 = 5 + 3t
-4 = 3t
t = -4/3

y: 4 + s = 7 + 6t
s = 3 + 6(-4/3) = 3 - 8 = -5

To verify plug into x.

x: 3 + 4s = -1 + 12t
3 + 4(-5) = -1 + 12(-4/3)
3 - 20 = -1 - 16
-17 = -17

There is a point of intersection.

x = 3 + 4s = 3 + 4(-5) = -17
y = 4 + s = 4 - 5 = -1
z = 1

The point of intersection is P(-17, -1, 1).
__________

Now find the plane containing the lines.

The normal vector n, of the plane is perpendicular to the directional vectors of both of the lines contained in the plane. Take the cross product.

n = <4, 1, 0> X <12, 6, 3> = <3, -12, 12>

Any non-zero multiple of n is also a normal vector to the plane. Divide by 3.

n = <1, -4, 4>

With the normal vector n and a point in the plane
P(-17, -1, 1) we can write the equation of the plane containing the two given lines.

1(x + 17) - 4(y + 1) + 4(z - 1) = 0
x + 17 - 4y - 4 + 4z - 4 = 0
x - 4y + 4z + 9 = 0

Answer 2

Equation Of Two Lines

Answer 3

First we make advantageous that lines could be located on one airplane. this occurs if the lines are parallel (which they are not) or in the event that they intersect. to detect factor of intersection, discover fee of t the place t = 3t, 2t = t, 3t = 8t ----> t = 0 -----> factor of intersection = (0, 0, 0) lines have direction vectors < a million, 2, 3 > and < 3, a million, 8 > Vector it is regularly occurring to airplane would be perpendicular to direction vectors of line. to detect vector it is perpendicular to those 2 vectors, merely take bypass product: < a million, 2, 3 > x < 3, a million, 8 > = < thirteen, a million, -5 > Now we are able to discover equation of airplane given a factor (0, 0, 0) and regularly occurring < thirteen, a million, -5 > thirteen (x - 0) + a million (y - 0) - 5 (z - 0) = 0 13x + y - 5z = 0

Answer 4

Thanks. This did really help