Cross Product Notes

Published on: September 29, 2026

This Cross Product Calculator multiplies two vectors into a third vector at right angles to both, and shows the working line by line. Each component of the answer comes from a two by two determinant built from the other two axes, which is what makes the result perpendicular. Swapping the order of the vectors reverses the answer. Following the same pattern every time keeps the alternating signs in order.

Step-by-step method

  1. Set up the problem.
  2. Use the cross product component formula.
  3. Substitute values and simplify.
See Example 1 Hide Example 1

Problem

\(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle,\; \vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Use the cross product component formula.
  3. Substitute values and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle\) and \(\vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\).

\(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle,\; \vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\)

Step 2

Step 2 - Use the cross product component formula.

In this problem: Each component of the cross product comes from a 2×2 determinant of the other two rows.

\(\vec{A} \times \vec{B} = \left\langle a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1} \right\rangle\)

Step 3

Step 3 - Substitute values and simplify.

In this problem: The cross product is \(\left\langle -3,\; 6,\; -3 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 26 - 35,\; 34 - 16,\; 15 - 24 \right\rangle = \left\langle -3,\; 6,\; -3 \right\rangle\)

Final Answer

\(\vec{A} \times \vec{B} = \left\langle -3,\; 6,\; -3 \right\rangle\)
See Example 2 Hide Example 2

Problem

\(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle,\; \vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Use the cross product component formula.
  3. Substitute values and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle\) and \(\vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\).

\(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle,\; \vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\)

Step 2

Step 2 - Use the cross product component formula.

In this problem: Each component of the cross product comes from a 2×2 determinant of the other two rows.

\(\vec{A} \times \vec{B} = \left\langle a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1} \right\rangle\)

Step 3

Step 3 - Substitute values and simplify.

In this problem: The cross product is \(\left\langle 0,\; 0,\; 1 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 00 - 01,\; 00 - 10,\; 11 - 00 \right\rangle = \left\langle 0,\; 0,\; 1 \right\rangle\)

Final Answer

\(\vec{A} \times \vec{B} = \left\langle 0,\; 0,\; 1 \right\rangle\)
See Example 3 Hide Example 3

Problem

\(\vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle,\; \vec{B} = \left\langle 1,\; 4,\; 2 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Use the cross product component formula.
  3. Substitute values and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle\) and \(\vec{B} = \left\langle 1,\; 4,\; 2 \right\rangle\).

\(\vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle,\; \vec{B} = \left\langle 1,\; 4,\; 2 \right\rangle\)

Step 2

Step 2 - Use the cross product component formula.

In this problem: Each component of the cross product comes from a 2×2 determinant of the other two rows.

\(\vec{A} \times \vec{B} = \left\langle a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1} \right\rangle\)

Step 3

Step 3 - Substitute values and simplify.

In this problem: The cross product is \(\left\langle -14,\; -1,\; 9 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle \left(-1\right)2 - 34,\; 31 - 22,\; 24 - \left(-1\right)1 \right\rangle = \left\langle -14,\; -1,\; 9 \right\rangle\)

Final Answer

\(\vec{A} \times \vec{B} = \left\langle -14,\; -1,\; 9 \right\rangle\)
See Example 4 Hide Example 4

Problem

\(\vec{A} = \left\langle 3,\; 0,\; 1 \right\rangle,\; \vec{B} = \left\langle 0,\; 2,\; -1 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Use the cross product component formula.
  3. Substitute values and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 3,\; 0,\; 1 \right\rangle\) and \(\vec{B} = \left\langle 0,\; 2,\; -1 \right\rangle\).

\(\vec{A} = \left\langle 3,\; 0,\; 1 \right\rangle,\; \vec{B} = \left\langle 0,\; 2,\; -1 \right\rangle\)

Step 2

Step 2 - Use the cross product component formula.

In this problem: Each component of the cross product comes from a 2×2 determinant of the other two rows.

\(\vec{A} \times \vec{B} = \left\langle a_{2}b_{3} - a_{3}b_{2},\; a_{3}b_{1} - a_{1}b_{3},\; a_{1}b_{2} - a_{2}b_{1} \right\rangle\)

Step 3

Step 3 - Substitute values and simplify.

In this problem: The cross product is \(\left\langle -2,\; 3,\; 6 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 0\left(-1\right) - 12,\; 10 - 3\left(-1\right),\; 32 - 00 \right\rangle = \left\langle -2,\; 3,\; 6 \right\rangle\)

Final Answer

\(\vec{A} \times \vec{B} = \left\langle -2,\; 3,\; 6 \right\rangle\)