Cross Product Calculator – a × b With Steps, Magnitude and Angle

Checked twice: perpendicular, and the angle measured two ways

A cross product makes two claims, and this page tests both. First, the result must be perpendicular to each parent — dotting it against a and against b must give exactly zero. Second, the angle between your vectors is measured from the dot product (cosine route) and from the cross magnitude (sine route): two independent computations that must agree.

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How it works, in 3 steps

How each answer is worked out

Deep dives

The unit vector: the direction, with the length divided out

Every cross product carries two pieces of information at once: a direction and a length. Sometimes only the direction matters — a surface normal for lighting, an axis to rotate about, a heading to point along. The unit vector is how you keep the direction and throw the length away. Take the cross product and divide every component by its own magnitude:

û = (a × b) ÷ |a × b|

For the vectors loaded above, a = (1, 2, 3) and b = (4, 5, 6), the cross product is (−3, 6, −3) and its magnitude is 7.3485. Dividing component by component gives the unit vector shown in the table, and the Advanced panel on this page computes it live for any pair you type in.

The check that this really is a unit vector is its own length: (−0.4082)² + (0.8165)² + (−0.4082)² comes to 1.0000 once rounding is allowed for, so û points exactly along a × b and stretches exactly one unit. The same division works on the inputs themselves. The unit vector along a is (0.2673, 0.5345, 0.8018), because |a| is 3.7417, and the unit vector along b is (0.4558, 0.5698, 0.6838), because |b| is 8.7750. Both appear in the Advanced panel beside the unit cross product, so you can compare all three directions at a glance.

Two cautions belong with the recipe. First, a zero cross product has no unit vector: dividing by a magnitude of zero is meaningless, which is the arithmetic echo of the geometry — parallel vectors fix no perpendicular direction at all. Second, in 2D mode the unit direction can only be (0, 0, 1) or (0, 0, −1), out of the page or into it, following the sign of the single z component. A positive signed area points toward you; a negative one points away.

Torque mode: the same product, read as turning

Torque is the cross product at work in the physical world, and it is the clearest reason the order of the two vectors matters. A lever arm r from a pivot to the point where a force F is applied produces a torque

τ = r × F

The torque vector points along the axle the turning happens about, following the right-hand rule from r toward F, and its magnitude is |r| |F| sin θ, where θ is the angle between arm and force. That sine is the whole story of leverage: a force at right angles to the arm turns with its full strength, a force along the arm does not turn at all, and a force at an angle contributes only its perpendicular part.

Take a lever arm r = (0.5, 0, 0) and a force F = (0, 40, 30). The cross product is (0 × 30 − 0 × 40, 0 × 0 − 0.5 × 30, 0.5 × 40 − 0 × 0) = (0, −15, 20), so the torque is 25 newton metres in magnitude: |r| is 0.5, |F| is 50, the force has no component along the arm so sin θ is 1, and 0.5 × 50 × 1 is 25. The torque points partly along negative y and partly along positive z because the force itself is split between those two directions — the cross product keeps that split, where a single-number answer would lose it.

The Advanced panel offers a torque reading for exactly this use: it reads your first vector as the lever arm and your second vector as the force, and reports the torque vector and its magnitude. It is a relabelling of the cross product you already computed, not a second calculation — the numbers are the same, and switching the reading on or off cannot change the main result. One honesty note: this calculator is unit-free, so the torque reading takes its units from whatever you typed. Metres for the arm and newtons for the force give newton metres; any other consistent pair of units gives the matching torque unit.

From area to surface normals: where the unit vector earns its keep

The magnitude of a cross product is an area, and the unit vector turns that area into an orientation — a combination computer graphics uses on every triangle it draws. A triangle in a 3D scene has two edge vectors, and crossing them gives a vector perpendicular to the triangle’s face. Divide by its length and you have the unit normal, the direction light calculations need: how bright a face looks depends on the angle between its normal and the light, which is a dot product, so the normal must carry direction only, with no area mixed in.

Take edges (2, 0, 0) and (0, 3, 0). Their cross product is (0 × 0 − 0 × 3, 0 × 0 − 2 × 0, 2 × 3 − 0 × 0) = (0, 0, 6). The parallelogram area is therefore 6, the triangle area is half of that, 3, and the unit normal is (0, 0, 1): straight out of the page. A second pair, edges (1, 0, 0) and (1, 1, 0), gives cross product (0, 0, 1), parallelogram area 1, triangle area 0.5, and the same unit normal (0, 0, 1). Same direction, different area — which is precisely why graphics keeps the two readings separate.

Order matters here exactly as it does for torque: swapping the two edges flips the normal to its opposite, and a renderer reading the flipped normal shades the face as if it faced the other way. The same unit normal also writes the plane through the origin in one line: with normal (−0.4082, 0.8165, −0.4082), the plane is −0.4082x + 0.8165y − 0.4082z = 0. Every point of the plane satisfies it, because dotting the normal with any vector lying in the plane gives zero — the perpendicularity check this page already performs, now doing geometry work.

The determinant, sign by sign: the cofactor pattern that explains the j flip

The determinant expansion is easier to trust once its sign pattern is out in the open. Expanding along the top row of the i, j, k determinant, the three columns carry the cofactor signs plus, minus, plus. The i component keeps its small determinant as it stands, the j component has its small determinant negated, and the k component keeps its own. That single minus sign on the middle column is the whole reason the j component is written a3b1 − a1b3, with the subtraction running in the opposite order from its neighbours.

Each small determinant follows the same two-by-two rule: for a grid with top row p, q and bottom row r, s, the determinant is ps − qr, the product of the falling diagonal minus the product of the rising diagonal. The working panels above show every one of these steps for your vectors, so a hand calculation can be compared line by line instead of only at the final answer.

Four mistakes account for most wrong answers, and each has a tell. Forgetting the j flip gives (−3, −6, −3) for the loaded vectors instead of (−3, 6, −3): only the middle sign is wrong, which is the fingerprint of this error. Swapping the two rows negates all three components, giving (3, −6, 3). Adding inside a small determinant instead of subtracting breaks the perpendicularity check — the result no longer dots to zero against its parents, which is how you catch it. And normalising the inputs before crossing quietly changes the answer’s meaning: the cross product of two unit vectors has magnitude sin θ, not the parallelogram area, so the area is lost unless you multiply |a| |b| back in. When a result looks wrong, dot it against both parents first: a genuine cross product always gives zero twice.

Check yourself: four pairs to work by hand

The fastest way to own this calculation is to work a few pairs by hand and then check them here. Each pair below is chosen to teach one thing: a routine case, an axis case where the answer is a single basis vector, a case with mixed signs, and the 2D version where the whole answer is one signed number. The dot-product column is the built-in check — a correct cross product is perpendicular to both parents, so both dots are zero in the 3D cases.

Verify the first pair against its parents: (1, −2, 6) dotted with (2, 1, 0) is 2 − 2 + 0 = 0, and dotted with (0, 3, 1) is 0 − 6 + 6 = 0. Verify the third pair the same way: (−6, 8, 13) dotted with (3, −1, 2) is −18 − 8 + 26 = 0, and dotted with (1, 4, −2) is −6 + 32 − 26 = 0. If either dot fails to vanish on your own working, the error is in one of the small determinants — and the sign table in the previous section says which one to recheck first.

One more measurement belongs to the same family and lives in the Advanced panel: add a third vector c and form the scalar triple product (a × b) · c, the signed volume of the box the three vectors span. For the loaded vectors, c = (0, 0, 1) gives (a × b) · c = −3, so the box volume is 3. The sign records orientation — swapping any two of the three vectors flips it — and a volume of zero means the three vectors lie flat in a single plane. Try c = (1, 1, 0) as well: the triple product is −3 + 6 = 3, volume 3 again, because only the part of c sticking out of the a–b plane contributes.

The scalar triple product: the volume of a parallelepiped

Three vectors define a solid, and one number gives its volume. The triple product is the cross product put to work a second time.

The scalar triple product of three vectors a, b, and c is a dot (b cross c), the dot product of the first vector with the cross product of the other two, and its absolute value is the volume of the parallelepiped, the slanted box, whose three edges from one corner are a, b, and c. It is computed as a determinant, with the three vectors as the rows or the columns, and the sign carries the orientation: a positive value means the three vectors form a right-handed set in the order given, and swapping any two vectors reverses the sign, exactly as swapping the two vectors of a cross product reverses its direction. For the vectors a = (1, 2, 3), b = (4, 5, 6), and c = (7, 8, 10), the triple product is negative 3, so the parallelepiped they define has a volume of 3 cubic units, and the sign is negative, so the set is left-handed in that order.

The zero case is as informative as the number. A triple product of zero means the three vectors lie in a single plane, coplanar, so the slanted box they define is flat and its volume is nothing: one of the three vectors can be built from the other two, and the three do not span three dimensions. The same test, applied to two vectors, is the angle test: the magnitude of the cross product is the product of the magnitudes times the sine of the angle between them, so a zero cross product means the sine is zero, the angle is 0 or 180 degrees, and the vectors are parallel, pointing the same way or opposite ways. Both tests are used wherever independence matters: in geometry, to test whether three points and a fourth define a volume or a plane; in mechanics, to test whether forces span the space they act in; and in computer graphics, to test whether a surface normal is well defined. The companion guide compares the dot product and the cross product, with the worked examples, so the triple product here can be read as the natural next step from the two products there.

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