Find the equation of the plane containing both lines...?

Question

Find the equation of the plane containing these two lines...

L1(t) = < -7, 3, -2 > + t < 4, 0, -1 >

L2(t) = < 0, 1, -7 > + t < 1, 2, 3 >

I'm so confused. Please help.

Answer 1

Find the equation of the plane containing these two lines.

L1(s) = < -7, 3, -2 > + s< 4, 0, -1 >
L2(t) = < 0, 1, -7 > + t< 1, 2, 3 >

The two lines are coplanar only if:
(i) They are parallel
(ii) They intersect

Otherwise they are skew lines and no plane can contain them both.

Since the directional vectors of the two lines are not non-zero multiples of each other they are not parallel. So we need to find a point of intersection or they are skew. If a point of intersection exists, the two lines are equal at that point. Set them equal.

x = -7 + 4s = t
y = 3 = 1 + 2t
z = -2 - s = -7 + 3t

Use the equation for y to solve for t.

3 = 1 + 2t
2 = 2t
t = 1

Plug into the equation for x to solve for s:

-7 + 4s = t
-7 + 4s = 1
4s = 8
s = 2

Plug into the equation for z to see if the solution is consistent.

-2 - s = -7 + 3t
-2 - 2 = -7 + 3
-4 = -4

The solution is consistent so there is a plane that contains both lines. The point of intersection of the two lines can be obtained from the equation of either plane. Let's use the first one.

x = -7 + 4s = -7 + 4*2 = 1
y = 3
z = -2 - s = -2 - 2 = -4

The point of intersection is P(1, 3, -4).

The normal vector n, is orthogonal to both directional vectors of the plane. Take the cross product.

n = < 4, 0, -1 > X <1, 2, 3> = <2, -13, 8>

With the normal vector of the plane and a point on the plane we can write the equation of the plane. Let's choose
P(1, 3, -4).

2(x - 1) - 13(y - 3) + 8(z + 4) = 0
2x - 2 - 13y + 39 + 8z + 32 = 0
2x - 13y + 8z + 69 = 0

Answer 2

the two strains can in common terms be interior the comparable airplane is they're parallel or in the event that they intersect. The directional vectors of the strains are: v1 = <2, one million, 0> v2 = <-one million, one million, one million> via fact the directional vectors at the instant are not non-0 multiples of one yet another they at the instant are not parallel. So verify to work out in the event that they have a factor of intersection. x = 2t = 3 - r y = 2 + t = 2 + r z = one million = r From z we see that: r = one million Plug into x and sparkling up for t. x = 2t = 3 - one million = 2 t = one million Plug the two into y to work out in the event that they answer is consistent. no remember if it is, there's a factor of intersection. y = 2 + one million = 2 + one million 3 = 3 the answer is consistent. there's a factor of intersection. %. the two line and plug in to discover the factor of intersection. L1: x = 2t = 2*one million = 2 y = 2 + t = 2 + one million = 3 z = one million the factor of intersection is P(2, 3, one million). the traditional vector n, of the airplane containing the two strains would be orthogonal to the directional vectors of the two strains. Take the bypass product. n = v1 X v2 = <2, one million, 0> X <-one million, one million, one million> = With the traditional vector n, to the airplane and a factor P, interior the airplane we can write the equation to the airplane. bear in suggestions, the traditional vector is orthogonal to any vector that lies interior the airplane. And the dot manufactured from orthogonal vectors is 0. define R(x,y,z) to be an arbitrary factor interior the airplane. Then vector PR lies interior the airplane. n • PR = 0 n • = 0 • = 0 one million(x - 2) - 2(y - 3) + 3(z - one million) = 0 x - 2 - 2y + 6 + 3z - 3 = 0 x - 2y + 3x + one million = 0

Answer 3

very confusing situation. research onto the search engines. that will help!