How do you deterime if two vectors lie in a parallel planes?

Question

If you have the vectors (x=4+5t, y=5+5t, z=1-4t) and (x=4+s, y=-6+8s, z=7-3s) how could you prove that they lie in parallel planes?

If you have the vectors (x=4+5t, y=5+5t, z=1-4t) and (x=4+s, y=-6+8s, z=7-3s) how could you prove that they lie in parallel planes?


I read the first response but the problem is I only have equations for the lines and in order to define a plane you need a line and a point not on that line.

Answer 1

two lines lie in parallel planes if the lines are skew...

to do that... verify that they are not intersecting...
then verify also that they are not parallel... (this one is easier as their direction vectors are not scalar multiples)

to show nonintersection...
solve s & t from x & y... but it will not fit z.
thus the two lines are non-intersecting.

to find the parallel planes...
get the cross-product of the two direction vectors...
〈 5 , 5, -4 〉 × 〈 1 , 8 , -3 〉
... the result is the normal vector for the parallel planes...

to get the planes... choose two different points from the two lines... one from each line.

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Answer 2

You didn't ask if the two vectors were parallel, but rather if they lie in parallel planes. Vectors have magnitude and direction but are not specifically located. So you can always find a plane that contains both vectors.

However, these are the equations of two lines, not just two vectors. And lines are specifically located. If the lines are parallel or they intersect, then there is always a plane that contains both of them. If not, then no single plane can contain both of the lines, but there are always two parallel planes that do.
____________

If the two directional vectors, v1 and v2, of the lines are not parallel, then neither are the lines.

v1 = <5, 5, -4>
v2 = <1, 8, -3>

Clearly, the vectors are not multiples of each other so the lines are not parallel.

If the lines intersect they lie in one plane. Check for a point of intersection of the two lines.

Set the parametric equations of the two planes equal to see if there is a consistent solution.

x = 4 + 5t = 4 + s
y = 5 + 5t = -6 + 8s
z = 1 - 4t = 7 - 3s

Solving we have:

x - y = -1 = 10 - 7s
7s = 11
s = 11/7

Plug back into the equation for x to solve for t.

x = 4 + 5t = 4 + s
5t = s = 11/7
t = 11/35

(s, t) = (11/7, 11/35)

Plug the values for s and t into the equations for z to see if there is a consistent solution.

z = 1 - 4t = 7 - 3s
3s = 6 + 4t
3(11/7) = 6 + 4(11/35)
33/7 ≠ 254/35
165/35 ≠ 254/35

There is no consistent solution so the lines do not intersect. Since they are also not parallel, that means they are skew. No one plane can contain them both.

Find two parallel planes that contain the lines. The parallel planes will have the same normal vector n. And it will be perpendicular to both of the directional vectors v1 and v2, of the lines since they lie in the two parallel planes to be determined. Take the cross product.

n = v1 X v2 = <5, 5, -4> X <1, 8, -3> = <17, 11, 35>

You can find a point in each plane by setting s and t equal to zero. With a point in each plane and the normal vector to the planes we can write the equations of the planes.

The first plane is:

17(x - 4) + 11(y - 5) + 35(z - 1) = 0
17x - 68 + 11y - 55 + 35z - 35 = 0
17x + 11y + 35z - 158 = 0

The second plane is:

17(x - 4) + 11(y + 6) + 35(z - 7) = 0
17x - 68 + 11y + 66 + 35z - 245 = 0
17x + 11y + 35z - 247 = 0

Answer 3

Well you provided equations for lines. But if two vectors are parallel, then the cross product of two said vectors equals zero

Your two vectors that are in the same direction as the line are

<5,5,-4> and <1,8,-3>

The cross product of the two vectors given is

<32-15, 15-4, 40-5>

<17, 11, 35>

As you can see, the vector cross product is not equal to zero or the zero vector <0,0,0>

So the two lines are not parralel because the two vectors are not parallel. And if two lines intersect, then the two planes containing each of the lines must also intersect, and are therefore not parallel.

Answer 4

How do you prove two planes are coincident?



First answer by Rdunlop. Last edit by Rdunlop. Contributor trust: 13 [recommend contributor]. Question popularity: 2 [recommend question]

Answer
By "coincident", do you mean that they lie on the same plane (i.e. they are instances of the same plane)? Assuming that's what you're after:

To begin with, you should have a plane equation defining each plane in the following form:

ax + by + cz + d = 0

This form can be re-written in vector form as:

(a,b,c) dot (x,y,z) + d = 0

Written in this form, (a,b,c) is a vector that defines the normal of the plane (a vector perpendicular to the plane), Which determines the orientation of the plane around the origin. The scalar value d defines a distance that the plane is offset from the origin, times the length of vector (a,b,c).

In case you aren't familiar with the dot product, it is a way of multiplying two vectors resulting in a scalar value that represents the angle and the magnitude of the two vectors. Specifically:

v1 dot v2 = |v1| |v2| cos(theta)

where:


theta is the angle between the vectors
|v| is the length of a vector
To compare the two planes, the first thing that we will want to do is "normalize" the plane equations, that is, scale them so that the length of (a,b,c) is 1.0. To do this, divide each constant in the equation by the length of the vector (a,b,c):

(ax + by + cz + d) / |(a, b, c)| = 0

The normalized version of the plane equation defines the same plane as the original plane equation. The difference is that now we can readily use the dot product to determine the angle between the two plane's normals, as "v1 dot v2 = cos(theta)" when |v1| and |v2| are both equal to 1.0. The cosine of zero degrees is 1.0, so if we calculate the dot product of the normal vectors of the two planes we can determine if the normals, and thus the planes, are parallel:

cos(theta) = (a1,b1,c1) dot (a2,b2,c2)
IF cos(theta) < 1.0
the planes are not parallel, and thus cannot be coincidental.
IF cos(theta) = 1.0
the planes are parallel, and may be coincidental
IF cos(theta) = -1.0
the planes are parallel, with normals facing opposite directions
the planes may be coincidental

If the planes are not parallel, you don't have to test any further, they are not coincidental. If they are parallel, then you can determine if they are coincidental by comparing the normalized d from each equation to see if they are equal, flipping the sign if the normals are facing opposite directions:

IF d1 = cos(theta) d2
the planes are coincident


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Answer 5

the traditional vector n, of the parallel planes containing the skew lines will be orthogonal to the directional vectors v1 and v2 of both lines. Take the flow product. n = v1 X v2 Then detect a element P and Q on each and each line. With the traditional vector of the parallel planes and some extent on each and each line (and hence in each and each airplane) the equations of both parallel planes containing both skew lines can somewhat be derived.

Answer 6

if the lines are parallel,then the angle between them is ZERO......thats all i remember...