Question

What is the equation of the planes containing two lines???

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

Hope it helps you buddy...
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