The Cross Product Calculator computes the cross product of two three-dimensional vectors. It also calculates the magnitude, unit vector direction, checks whether vectors are parallel, and computes the dot product.
Enter the components of two 3D vectors to get instant results with step-by-step solutions showing the determinant formula expansion.
Your calculations will appear here
The cross product (also called the vector product) of two three-dimensional vectors a and b produces a new vector that is perpendicular to both a and b. It is defined as a x b = |a||b|sin(theta) n, where theta is the angle between the vectors and n is the unit normal determined by the right-hand rule.
The most practical way to compute the cross product is through the determinant of a 3x3 matrix whose first row contains the unit vectors i, j, k and whose second and third rows contain the components of a and b respectively. Expanding along the first row gives: a x b = (a_y * b_z - a_z * b_y) i - (a_x * b_z - a_z * b_x) j + (a_x * b_y - a_y * b_x) k.
The magnitude of the cross product |a x b| equals the area of the parallelogram formed by the two vectors. Dividing by two gives the area of the triangle they span. This geometric interpretation makes the cross product essential in computational geometry and computer graphics for calculating surface areas and determining face orientations.
Key algebraic properties distinguish the cross product from the dot product. The cross product is anti-commutative: a x b = -(b x a). It is distributive over addition: a x (b + c) = a x b + a x c. However, it is NOT associative: a x (b x c) does not generally equal (a x b) x c. The cross product of any vector with itself is the zero vector, and the cross product of parallel vectors is always zero.
In physics, the cross product appears in torque (tau = r x F, where r is the position vector and F is the force), angular momentum (L = r x p), and the magnetic force on a charged particle (F = qv x B). In computer graphics, cross products compute surface normals for lighting calculations, determine the winding order of polygon vertices, and test whether a point lies inside a triangle.
Problem: Find the cross product of i = (1, 0, 0) and j = (0, 1, 0).
Solution: Apply the formula: i component = (0)(0) - (0)(1) = 0. j component = -((1)(0) - (0)(0)) = 0. k component = (1)(1) - (0)(0) = 1. So i x j = (0, 0, 1) = k.
Answer: (0, 0, 1)
Problem: Compute (2, 3, 4) x (5, 6, 7).
Solution: i component: (3)(7) - (4)(6) = 21 - 24 = -3. j component: -((2)(7) - (4)(5)) = -(14 - 20) = 6. k component: (2)(6) - (3)(5) = 12 - 15 = -3.
Answer: (-3, 6, -3)
Problem: Compute (2, 4, 6) x (1, 2, 3).
Solution: Note that (2, 4, 6) = 2 * (1, 2, 3), so the vectors are parallel. i component: (4)(3) - (6)(2) = 12 - 12 = 0. j component: -((2)(3) - (6)(1)) = -(6 - 6) = 0. k component: (2)(2) - (4)(1) = 4 - 4 = 0.
Answer: (0, 0, 0) -- the zero vector, confirming the vectors are parallel
Problem: Find the area of the triangle with vertices A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1).
Solution: Form edge vectors: AB = B - A = (-1, 1, 0), AC = C - A = (-1, 0, 1). Cross product: AB x AC = ((1)(1) - (0)(0), -((-1)(1) - (0)(-1)), (-1)(0) - (1)(-1)) = (1, 1, 1). Magnitude: sqrt(1 + 1 + 1) = sqrt(3). Triangle area = |AB x AC| / 2 = sqrt(3) / 2.
Answer: sqrt(3) / 2 (approximately 0.866)
Problem: A force F = (0, 0, 10) N is applied at position r = (3, 0, 0) m from the pivot. Find the torque.
Solution: Torque tau = r x F. i component: (0)(10) - (0)(0) = 0. j component: -((3)(10) - (0)(0)) = -30. k component: (3)(0) - (0)(0) = 0. So tau = (0, -30, 0) N*m.
Answer: (0, -30, 0) N*m -- the torque acts in the negative y-direction
Problem: Find a normal vector to the plane containing the direction vectors u = (1, 2, 0) and v = (3, 0, 1).
Solution: The cross product u x v is perpendicular to both u and v, hence normal to the plane. i component: (2)(1) - (0)(0) = 2. j component: -((1)(1) - (0)(3)) = -1. k component: (1)(0) - (2)(3) = -6.
Answer: (2, -1, -6)
Problem: Find the unit normal vector for the cross product result (2, -1, -6).
Solution: Magnitude = sqrt(4 + 1 + 36) = sqrt(41). Unit vector = (2/sqrt(41), -1/sqrt(41), -6/sqrt(41)).
Answer: (0.3123, -0.1562, -0.9370) approximately
Problem: A particle with charge q = 2 C moves with velocity v = (1, 0, 0) m/s through magnetic field B = (0, 1, 0) T. Find the force.
Solution: F = qv x B. First compute v x B: i = (0)(0) - (0)(1) = 0, j = -((1)(0) - (0)(0)) = 0, k = (1)(1) - (0)(0) = 1. So v x B = (0, 0, 1). Then F = 2 * (0, 0, 1) = (0, 0, 2) N.
Answer: (0, 0, 2) N -- the force is perpendicular to both velocity and field
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