What is a formula to get a vector perpendicular to another vector?

Question

What is a formula to get a three dimensional vector B lying on the plane perpendicular to a vector A?

That is, given a vector A, what is a formula f(angle,modulus) which gives a vector that is perpendicular to A, with said modulus and rotated through an angle?

Solution 1

If the two vectors are perpendicular then their dot product is zero.

So: v1(x1, y1, z1), v2(x2, y2, z2).

=> x1 * x2 + y1 * y2 + z1 * z2 = 0

You know (x1, y1, z1). Put arbitrary x2 andy2 and you will receive the corresponding z2:

z1 * z2 = -x1 * x2 - y1 * y2
=> z2 = (-x1 * x2 - y1 * y2) / z1

Be aware if z1 is 0. Then you are in the plane.

Solution 2

function (a,b,c)
{
    return (-b,a,0)
}

But this answer is not numerical stable when a,b are close to 0.

To avoid that case, use:

function (a,b,c) 
{
    return  c<a  ? (b,-a,0) : (0,-c,b) 
}

The above answer is numerical stable, because in case c < a then max(a,b) = max(a,b,c), then vector(b,-a,0).length() > max(a,b) = max(a,b,c) , and since max(a,b,c) should not be close to zero, so is the vector. The c > a case is similar.

Solution 3

Calculate the cross product AxC with another vector C which is not collinear with A.

There are many possible directions in the plane perpendicular to A. If you don't really care, which one to pick, just create an arbitrary vector C not collinear with A:

if (A2 != 0 || A3 != 0)
    C = (1, 0, 0);
else
    C = (0, 1, 0);
B = A x C; 

Solution 4

I believe that this should produce an arbitrary vector that is perpendicular to the given vector vec while remaining numerically stable regardless of the angle of vec (assuming that the magnitude of vec is not close to zero). Assume that Vec3D is a three dimensional vector of arbitrary numerical type.

Vec3D arbitrary_orthogonal(Vec3D vec)
{
  bool b0 = (vec[0] <  vec[1]) && (vec[0] <  vec[2]);
  bool b1 = (vec[1] <= vec[0]) && (vec[1] <  vec[2]);
  bool b2 = (vec[2] <= vec[0]) && (vec[2] <= vec[1]);

  return cross(vec, Vec3D(int(b0), int(b1), int(b2)));
}

Informal explanation

Exactly 1 and only 1 of the bools get set; bN gets set if dimension N has magnitude strictly less than all subsequent dimensions and not greater than all previous dimensions. We then have a unit vector with a single non-zero dimension that corresponds to a dimension of minimum magnitude in vec. The cross product of this with vec is orthogonal to vec by defn of cross product. Consider now that the cross product is numerically unstable only when the two vectors are very closely aligned. Consider that our unit vector is large in only a single dimension and that that dimension corresponds to the dimension where vec was small. It's thus guaranteed to be loosely orthogonal to vec before taking the cross product, with least orthogonality in the case where all dimensions of vec are equal. In this least-orthogonal case, we're still quite orthogonal given that our unit vector has all but one dimension 0 whereas vec has all equal. We thus avoid the unstable case of taking the cross product of two nearly-aligned vectors.

Solution 5

One way would be to find a rotation transform from the positive z-axis (or any other axis) to your given vector. Then transform <modulus * cos(angle), modulus * sin(angle), 0> using this transform.

def getPerpendicular(v1,modulus,angle):
    v2 = vector(0,0,1)
    v1_len = v2.length()

    axis = v1.cross_product(v2)
    sinAngle = axis.length() / v1_len       # |u x v| = |u| * |v| * sin(angle)
    cosAngle = v1.dot_product(v2) / v1_len  # u . v = |u| * |v| * cos(angle)
    axis = axis.normalize()
    # atan2(sin(a), cos(a)) = a, -pi < a < pi
    angle = math.atan2(sinAngle, cosAngle)

    rotationMatrix = fromAxisAngle(axis, angle)

    # perpendicular to v2
    v3 = vector(modulus*cos(angle),modulus*sin(angle),0)

    return rotationMatrix.multiply(v3);

To calculate the rotation matrix, see this article: WP: Rotation matrix from axis and angle

Another method would be to use quaternion rotation. It's a little more to wrap your head around, but it's less numbers to keep track of.

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