Find an equation for the plane that contains both parallel lines r1 =<2+2t,4+6t,

Question

Find an equation for the plane that contains both parallel lines r1 =<2+2t,4+6t,4t>andr2 =<42t,26t,2+4t>(Hint: since they are parallel, you only have one direction vector of the plane so far–find another so you can cross them!). Find an equation for the plane that contains both parallel lines r1 =<2+2t,4+6t,4t>andr2 =<42t,26t,2+4t>(Hint: since they are parallel, you only have one direction vector of the plane so far–find another so you can cross them!). Find an equation for the plane that contains both parallel lines r1 =<2+2t,4+6t,4t>andr2 =<42t,26t,2+4t>(Hint: since they are parallel, you only have one direction vector of the plane so far–find another so you can cross them!).

Explanation / Answer

Solution:

Given:

r1 =<2+2t,4+6t,4t>and r2 =<42t,26t,2+4t>

r1 = <2,-4,0> + t<2,6,-4>
r2 = <4,2,2> + s<-2,-6,4>

(-1) * <2,6,-4> = <-2,-6,4>

So the lines are parallel.
One directional vector of the plane is:
u = <-2, -6, 4>

Another one can be found by picking a point on each line.

Let t = 0 and the two points are:

P(2, -4, 0) and Q(4, 2, 2).

v = PQ = <Q - P> = <2, 6, 2>

The normal vector n, of the plane is perpendicular to both directional vectors.

Take the cross product.

n = u X v = <-2, -6, 4> X<2,6,2> = <-36 ,12,0>

Any non-zero multiple of n is also a normal vector to the plane. Divide by 12.

n = <-3, 1, 0>

With one point and the normal vector we can write the equation of the plane.

Let's choose P(2, -4, 0).

Remember, any vector that lies in the plane will be perpendicular to the normal vector and the product of normal vectors is zero.

n • <x - 2, y +4, z - 0> = <-3, 1, 0> • <x - 2, y + 4, z - 0>

2(x - 2) + 5(y +4 ) + 0 (z ) = 0
2x -4 + 5y +20 =0

2x +5y = -16

this the  plane that contains both parallel lines

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