How to find the vector parallel to two planes?

Answer 1

A vector can't be parallel to a plane, since a plane has not a single direction, as a vector does. You can find a vector in the same direction as the intersection of two planes. That vector will be perpendicular to the normal vectors of both planes, and this will be true no matter if the planes intersect or not.

The explanation to follow is for planes and vectors in R3. In a general vector space all of these formulas still hold, with the obvious modifications.

Plane ax + by + cz = g has normal vector (a,b,c), and likewise dx + ey + fz = h has normal vector (d,e,f). Now find a vector (u,v,w) perpendicular to both of these, by setting

(a,b,c) . (u,v,w) = 0
and
(d,e,f) . (u,v,w) = 0.

These can be combined by solving the matrix equation

[a b c]
[d e f] . (u,v,w) = 0.

You will get a family of vectors in a particular direction as your solution. Any member of this family will be parallel to the direction of the intersection of the two planes, if any, and will be in a skew direction to the two planes, if they are parallel. In this latter case you will get a two-parameter family of vectors as a solution, of course.

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

A vector parallel to two planes would be a vector that lies in both planes.

If we are talking about 3-space

Case 1. The planes are not parallel.
Take the cross product of the normals of both planes. The resultant will lie in both planes (i.e. it will be parallel to both planes).

Case 2. The two planes are parallel.
In this case take the normal vector for either plane and find a vector whose dot product with the normal plane is zero.

Answer 4

why two planes? The equation of a plane is in the form of Ax + By + Cz + D = 0

the vector normal to that plane is N =

now to find a vector parallel to the plane you need to find a vector perpendicular to the normal vector.

you can do this by finding a vector that has a zero dot product when dotted with the normal vector.

Answer 5

There are two different approaches:

In eucledean space the plane may be defined by normal vector to it (Nx, Ny, Nz). Equation of plane in this case is
x Nx + y Ny + z Nz = const.
If you are given two normals N1 and N2 to the two planes you can simply take vector product N1 x N2 = V and vector V will be parallel to both planes.

But the wording of your problem suggests that you actually operate in affine space and distances might not be defined. In this case you should find the intersection of the planes. Any straight line parallel to this intersection line will be parallel to both planes.

Answer 6

e.g, we have plane equation as;
ax+by+cz=d
now, vector perpendicular to plane is and vector parallel to the plane is ;