Cross Product Calculator

Enter the vector components to calculate their cross product (a × b) and see the step-by-step solution.

Vector a
⟨ , , ⟩
Vector b
⟨ , , ⟩

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How to Use the Calculator

This online cross product solver is an advanced mathematical tool designed to operate with vectors in three-dimensional space. Unlike the dot product, this operation generates a new vector that is orthogonal (perpendicular) to the original two.

How to enter your data:

  • Direct entry in R3: enter the three rectangular components (x, y, z) for each of the vectors.
  • Supported numerical formats: the input fields are flexible; you can use integers, decimals, exact fractions, and even irrational numbers (such as square roots). The algebraic engine will process these values symbolically to ensure there are no rounding errors during the calculation.

Result structure:

When processing your exercise, the tool provides an analytical solution, divided into two main sections:

  1. Quick answer and notations: in the top highlighted box, you will see the answer to your problem. The calculator will show the new resulting vector expressed in two standard ways: in component form (x, y, z) and as a linear combination of standard unit vectors (i, j, k). Additionally, this box will give you the magnitude of the resulting vector.
  2. Step-by-step solution: in the lower section, you will find the complete breakdown of the method used. The solver will show you how to set up the 3x3 matrix determinant, placing the unit vectors i, j, and k in the first row, and your vector components in the rows below. Then, you will see the cofactor expansion (solving the 2x2 minor determinants for each term), reducing the expressions term by term until the final vector is assembled.

Solved Exercises

The following are examples of problems solved by the calculator.

Calculate the cross product of the vectors \(\vec{a}=\langle 3, -2, 4\rangle\) and \(\vec{b}=\langle 1, 5, -3\rangle.\)

Result

The cross product of the given vectors is:

$$ \displaystyle \vec{a} \times \vec{b} = \langle -14, 13, 17 \rangle $$

Unit vector notation: $$ \displaystyle \vec{a} \times \vec{b} = -14\hat{i} + 13\hat{j} + 17\hat{k} $$

Magnitude of the resulting vector: $$ \displaystyle |\vec{a} \times \vec{b}| = \sqrt{654} \approx 25.57 $$

Step-by-step solution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \langle 3, -2, 4 \rangle \\[1em] \vec{b} = \langle 1, 5, -3 \rangle $$

2. Set up the general determinant.

We express the cross product using a 3x3 determinant, where the first row contains the unit vectors i, j, k, and the following rows contain the vector components.

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] a_x & a_y & a_z \\[0.5em] b_x & b_y & b_z \end{vmatrix} $$

3. Substitute the vector components.

We plug the entered vector values into the determinant while keeping their order (since the cross product is not commutative).

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.5em] 3 & -2 & 4 \\[0.5em] 1 & 5 & -3 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the main determinant into the unit components i, j, and k, each multiplied by their respective 2x2 minor determinants. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -2 & 4 \\[0.5em] 5 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 4 \\[0.5em] 1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\[0.5em] 1 & 5 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products to obtain each component separately.

First term (vector i):

$$ \displaystyle \hat{i} \begin{vmatrix} -2 & 4 \\[0.5em] 5 & -3 \end{vmatrix} = \hat{i} [(-2)(-3) - (4)(5)] \\[1em] = \hat{i} [6 - 20] \\[1em] = -14\hat{i} $$

Second term (vector j):

It is important not to forget the negative sign on this term.

$$ \displaystyle - \hat{j} \begin{vmatrix} 3 & 4 \\[0.5em] 1 & -3 \end{vmatrix} = - \hat{j} [(3)(-3) - (4)(1)] \\[1em] = - \hat{j} [-9 - 4] \\[1em] = 13\hat{j} $$

Third term (vector k):

$$ \displaystyle \hat{k} \begin{vmatrix} 3 & -2 \\[0.5em] 1 & 5 \end{vmatrix} = \hat{k} [(3)(5) - (-2)(1)] \\[1em] = \hat{k} [15 - (-2)] \\[1em] = 17\hat{k} $$

Assemble the vector

Now that we have all three components, we can put together the resulting vector:

$$ \displaystyle \vec{a} \times \vec{b} = -14\hat{i} + 13\hat{j} + 17\hat{k} \\[1em] \vec{a} \times \vec{b} = \langle -14, 13, 17 \rangle $$
Find the cross product of the vectors \(\langle 0, 2, 0\rangle\) and \(\langle -4, 0, 1\rangle.\)

Result

The cross product of the given vectors is:

$$ \displaystyle \vec{a} \times \vec{b} = \langle 2, 0, 8 \rangle $$

Unit vector notation: $$ \displaystyle \vec{a} \times \vec{b} = 2\hat{i} + 8\hat{k} $$

Magnitude of the resulting vector: $$ \displaystyle |\vec{a} \times \vec{b}| = 2 \sqrt{17} \approx 8.25 $$

Step-by-step solution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \langle 0, 2, 0 \rangle \\[1em] \vec{b} = \langle -4, 0, 1 \rangle $$

2. Set up the general determinant.

We express the cross product using a 3x3 determinant, where the first row contains the unit vectors i, j, k, and the following rows contain the vector components.

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] a_x & a_y & a_z \\[0.5em] b_x & b_y & b_z \end{vmatrix} $$

3. Substitute the vector components.

We plug the entered vector values into the determinant while keeping their order (since the cross product is not commutative).

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.5em] 0 & 2 & 0 \\[0.5em] -4 & 0 & 1 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the main determinant into the unit components i, j, and k, each multiplied by their respective 2x2 minor determinants. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 2 & 0 \\[0.5em] 0 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & 0 \\[0.5em] -4 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\[0.5em] -4 & 0 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products to obtain each component separately.

First term (vector i):

$$ \displaystyle \hat{i} \begin{vmatrix} 2 & 0 \\[0.5em] 0 & 1 \end{vmatrix} = \hat{i} [(2)(1) - (0)(0)] \\[1em] = \hat{i} [2 - 0] \\[1em] = 2\hat{i} $$

Second term (vector j):

It is important not to forget the negative sign on this term.

$$ \displaystyle - \hat{j} \begin{vmatrix} 0 & 0 \\[0.5em] -4 & 1 \end{vmatrix} = - \hat{j} [(0)(1) - (0)(-4)] \\[1em] = - \hat{j} [0 - 0] \\[1em] = 0\hat{j} $$

Third term (vector k):

$$ \displaystyle \hat{k} \begin{vmatrix} 0 & 2 \\[0.5em] -4 & 0 \end{vmatrix} = \hat{k} [(0)(0) - (2)(-4)] \\[1em] = \hat{k} [0 - (-8)] \\[1em] = 8\hat{k} $$

Assemble the vector

Now that we have all three components, we can put together the resulting vector:

$$ \displaystyle \vec{a} \times \vec{b} = 2\hat{i} + 8\hat{k} \\[1em] \vec{a} \times \vec{b} = \langle 2, 0, 8 \rangle $$
Obtain the product a × b of the vectors \(\vec{a}=\langle 1/2, 3, -1\rangle\) and \(\vec{b}=\langle 2, -1/3, 4 \rangle.\)

Result

The cross product of the given vectors is:

$$ \displaystyle \vec{a} \times \vec{b} = \left\langle \frac{35}{3}, -4, -\frac{37}{6} \right\rangle \approx \langle 11.67, -4, -6.17 \rangle $$

Unit vector notation: $$ \displaystyle \vec{a} \times \vec{b} = \frac{35}{3}\hat{i} - 4\hat{j} - \frac{37}{6}\hat{k} $$

Magnitude of the resulting vector: $$ \displaystyle |\vec{a} \times \vec{b}| = \frac{37 \sqrt{5}}{6} \approx 13.79 $$

Step-by-step solution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \left\langle \frac{1}{2}, 3, -1 \right\rangle \\[1em] \vec{b} = \left\langle 2, -\frac{1}{3}, 4 \right\rangle $$

2. Set up the general determinant.

We express the cross product using a 3x3 determinant, where the first row contains the unit vectors i, j, k, and the following rows contain the vector components.

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] a_x & a_y & a_z \\[0.5em] b_x & b_y & b_z \end{vmatrix} $$

3. Substitute the vector components.

We plug the entered vector values into the determinant while keeping their order (since the cross product is not commutative).

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.5em] \frac{1}{2} & 3 & -1 \\[0.5em] 2 & -\frac{1}{3} & 4 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the main determinant into the unit components i, j, and k, each multiplied by their respective 2x2 minor determinants. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 3 & -1 \\[0.5em] -\frac{1}{3} & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} \frac{1}{2} & -1 \\[0.5em] 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} \frac{1}{2} & 3 \\[0.5em] 2 & -\frac{1}{3} \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products to obtain each component separately.

First term (vector i):

$$ \displaystyle \hat{i} \begin{vmatrix} 3 & -1 \\[0.5em] -\frac{1}{3} & 4 \end{vmatrix} = \hat{i} \left[(3)(4) - (-1)\left(-\frac{1}{3}\right)\right] \\[1em] = \hat{i} \left[12 - \frac{1}{3}\right] \\[1em] = \frac{35}{3}\hat{i} $$

Second term (vector j):

It is important not to forget the negative sign on this term.

$$ \displaystyle - \hat{j} \begin{vmatrix} \frac{1}{2} & -1 \\[0.5em] 2 & 4 \end{vmatrix} = - \hat{j} \left[\left(\frac{1}{2}\right)(4) - (-1)(2)\right] \\[1em] = - \hat{j} [2 - (-2)] \\[1em] = -4\hat{j} $$

Third term (vector k):

$$ \displaystyle \hat{k} \begin{vmatrix} \frac{1}{2} & 3 \\[0.5em] 2 & -\frac{1}{3} \end{vmatrix} = \hat{k} \left[\left(\frac{1}{2}\right)\left(-\frac{1}{3}\right) - (3)(2)\right] \\[1em] = \hat{k} \left[-\frac{1}{6} - 6\right] \\[1em] = -\frac{37}{6}\hat{k} $$

Assemble the vector

Now that we have all three components, we can put together the resulting vector:

$$ \displaystyle \vec{a} \times \vec{b} = \frac{35}{3}\hat{i} - 4\hat{j} - \frac{37}{6}\hat{k} \\[1em] \vec{a} \times \vec{b} = \left\langle \frac{35}{3}, -4, -\frac{37}{6} \right\rangle \approx \langle 11.67, -4, -6.17 \rangle $$
Calculate the cross multiplication of the vectors with decimal components \(\langle -1.5, 2.4, 0\rangle\) and \(\langle 3.2, -1, 2.5\rangle.\)

Result

The cross product of the given vectors is:

$$ \displaystyle \vec{a} \times \vec{b} = \left\langle 6, \frac{15}{4}, -\frac{309}{50} \right\rangle \approx \langle 6, 3.75, -6.18 \rangle $$

Unit vector notation: $$ \displaystyle \vec{a} \times \vec{b} = 6\hat{i} + \frac{15}{4}\hat{j} - \frac{309}{50}\hat{k} $$

Magnitude of the resulting vector: $$ \displaystyle |\vec{a} \times \vec{b}| = \frac{3 \sqrt{3} \sqrt{32687}}{100} \approx 9.39 $$

Step-by-step solution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \langle -1.5, 2.4, 0 \rangle = \left\langle -\frac{3}{2}, \frac{12}{5}, 0 \right\rangle \\[1em] \vec{b} = \langle 3.2, -1, 2.5 \rangle = \left\langle \frac{16}{5}, -1, \frac{5}{2} \right\rangle $$

2. Set up the general determinant.

We express the cross product using a 3x3 determinant, where the first row contains the unit vectors i, j, k, and the following rows contain the vector components.

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] a_x & a_y & a_z \\[0.5em] b_x & b_y & b_z \end{vmatrix} $$

3. Substitute the vector components.

We plug the entered vector values into the determinant while keeping their order (since the cross product is not commutative).

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.5em] -\frac{3}{2} & \frac{12}{5} & 0 \\[0.5em] \frac{16}{5} & -1 & \frac{5}{2} \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the main determinant into the unit components i, j, and k, each multiplied by their respective 2x2 minor determinants. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} \frac{12}{5} & 0 \\[0.5em] -1 & \frac{5}{2} \end{vmatrix} - \hat{j} \begin{vmatrix} -\frac{3}{2} & 0 \\[0.5em] \frac{16}{5} & \frac{5}{2} \end{vmatrix} + \hat{k} \begin{vmatrix} -\frac{3}{2} & \frac{12}{5} \\[0.5em] \frac{16}{5} & -1 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products to obtain each component separately.

First term (vector i):

$$ \displaystyle \hat{i} \begin{vmatrix} \frac{12}{5} & 0 \\[0.5em] -1 & \frac{5}{2} \end{vmatrix} = \hat{i} \left[\left(\frac{12}{5}\right)\left(\frac{5}{2}\right) - (0)(-1)\right] \\[1em] = \hat{i} [6 - 0] \\[1em] = 6\hat{i} $$

Second term (vector j):

It is important not to forget the negative sign on this term.

$$ \displaystyle - \hat{j} \begin{vmatrix} -\frac{3}{2} & 0 \\[0.5em] \frac{16}{5} & \frac{5}{2} \end{vmatrix} = - \hat{j} \left[\left(-\frac{3}{2}\right)\left(\frac{5}{2}\right) - (0)\left(\frac{16}{5}\right)\right] \\[1em] = - \hat{j} \left[-\frac{15}{4} - 0\right] \\[1em] = \frac{15}{4}\hat{j} $$

Third term (vector k):

$$ \displaystyle \hat{k} \begin{vmatrix} -\frac{3}{2} & \frac{12}{5} \\[0.5em] \frac{16}{5} & -1 \end{vmatrix} = \hat{k} \left[\left(-\frac{3}{2}\right)(-1) - \left(\frac{12}{5}\right)\left(\frac{16}{5}\right)\right] \\[1em] = \hat{k} \left[\frac{3}{2} - \frac{192}{25}\right] \\[1em] = -\frac{309}{50}\hat{k} $$

Assemble the vector

Now that we have all three components, we can put together the resulting vector:

$$ \displaystyle \vec{a} \times \vec{b} = 6\hat{i} + \frac{15}{4}\hat{j} - \frac{309}{50}\hat{k} \\[1em] \vec{a} \times \vec{b} = \left\langle 6, \frac{15}{4}, -\frac{309}{50} \right\rangle \approx \langle 6, 3.75, -6.18 \rangle $$
Find the result of the cross product \(\langle 1, 1, 1\rangle \times \langle -1, -1, -1\rangle.\)

Result

The cross product of the given vectors is:

$$ \displaystyle \vec{a} \times \vec{b} = \langle 0, 0, 0 \rangle $$

Unit vector notation: $$ \displaystyle \vec{a} \times \vec{b} = 0\hat{i} + 0\hat{j} + 0\hat{k} $$

Magnitude of the resulting vector: $$ \displaystyle |\vec{a} \times \vec{b}| = 0 $$

Step-by-step solution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \langle 1, 1, 1 \rangle \\[1em] \vec{b} = \langle -1, -1, -1 \rangle $$

2. Set up the general determinant.

We express the cross product using a 3x3 determinant, where the first row contains the unit vectors i, j, k, and the following rows contain the vector components.

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] a_x & a_y & a_z \\[0.5em] b_x & b_y & b_z \end{vmatrix} $$

3. Substitute the vector components.

We plug the entered vector values into the determinant while keeping their order (since the cross product is not commutative).

$$ \displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.5em] 1 & 1 & 1 \\[0.5em] -1 & -1 & -1 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the main determinant into the unit components i, j, and k, each multiplied by their respective 2x2 minor determinants. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products to obtain each component separately.

First term (vector i):

$$ \displaystyle \hat{i} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} = \hat{i} [(1)(-1) - (1)(-1)] \\[1em] = \hat{i} [-1 - (-1)] \\[1em] = 0\hat{i} $$

Second term (vector j):

It is important not to forget the negative sign on this term.

$$ \displaystyle - \hat{j} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} = - \hat{j} [(1)(-1) - (1)(-1)] \\[1em] = - \hat{j} [-1 - (-1)] \\[1em] = 0\hat{j} $$

Third term (vector k):

$$ \displaystyle \hat{k} \begin{vmatrix} 1 & 1 \\[0.5em] -1 & -1 \end{vmatrix} = \hat{k} [(1)(-1) - (1)(-1)] \\[1em] = \hat{k} [-1 - (-1)] \\[1em] = 0\hat{k} $$

Assemble the vector

Now that we have all three components, we can put together the resulting vector:

$$ \displaystyle \vec{a} \times \vec{b} = 0\hat{i} + 0\hat{j} + 0\hat{k} \\[1em] \vec{a} \times \vec{b} = \langle 0, 0, 0 \rangle $$

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Daniel Machado

Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.

Last updated: (2 weeks ago)