# Cross Product

## Right-Hand Rule

A triple of non-coplanar vectors a, b and c are said to form a right-handed coordinate system if, being reduced to a common origin, these vectors are located, respectively, as the index finger, middle finger and the thumb of the right hand.

## Definition of the Cross Product

The cross product of vector a and vector b is called vector c, denoted by c = a × b and satisfying the following requirements:

1. The length of vector c is equal to the product of the lengths of vectors a and b and the sine of the angle φ between them, that is
$\Vert{\mathbf{c}}\Vert = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert \sin\varphi$
2. Vector c is orthogonal to each of the vectors a and b;
3. Vector c is directed so that the three vectors a, b and c is right-handed.

## Some Properties of the Cross Product

The absolute value of the cross product of a and b is equal to the area of the parallelogram spanned by these vectors:

$A = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert \sin\varphi$

The angle between two vectors can be expressed in terms of their cross product:

$\sin\varphi = \frac{\Vert{\mathbf{a} \times \mathbf{b}}\Vert}{\Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert}$

The cross product is anticommutative:

$\mathbf{a} \times \mathbf{b} = - \left({\mathbf{a} \times \mathbf{b}}\right)$

The cross product is associative only with respect to scalar multiplication:

$\left({\lambda\mathbf{a}}\right) \times \left({\mu\mathbf{b}}\right) = \lambda\mu\left({\mathbf{a} \times \mathbf{b}}\right),$

where $$\lambda, \mu$$ are real numbers.

Distributive property of the cross product over addition of vectors:

$\mathbf{a} \times \left({\mathbf{b} + \mathbf{c}}\right) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$

The cross product of vectors a and b is the zero vector if a and b are parallel (collinear):

$\mathbf{a} \times \mathbf{b} = \mathbf{0} \text{ if } \mathbf{a} \parallel \mathbf{b} \left({\varphi = 0}\right)$

Cross product of the unit vectors:

$\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}$

Cross product of distinct unit vectors:

$\mathbf{i} \times \mathbf{j} = \mathbf{k}, \;\mathbf{j} \times \mathbf{k} = \mathbf{i},\; \mathbf{k} \times \mathbf{i} = \mathbf{j}$

## Cross Product in Coordinate Form

If $$\mathbf{a} = \left({X_1, Y_1, Z_1}\right)$$ and $$\mathbf{b} = \left({X_2, Y_2, Z_2}\right),$$ then the cross product of these vectors is given by

$\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{Y_1}} & {{Z_1}}\\ {{Y_2}} & {{Z_2}} \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{c}} {{X_1}} & {{Z_1}}\\ {{X_2}} & {{Z_2}} \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{c}} {{X_1}} & {{Y_1}}\\ {{X_2}} & {{Y_2}} \end{array}} \right|\mathbf{k}$

## Moment of a Force (Torque)

If vector $$\mathbf{F_B}$$ represents the force applied at some point B and vector $$\mathbf{r_{AB}}$$ goes from some point A to point B, then the cross product $$\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F_B}$$ represents the moment of force $$\mathbf{F_B}$$ relative to point A.

$\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F_B}$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Given vectors $$\mathbf{a}\left({1,2,3}\right)$$ and $$\mathbf{b}\left({2,4,7}\right).$$ Find the coordinates of the cross product $$\mathbf{a} \times \mathbf{b}.$$

### Example 2

Force $$\mathbf{F}\left({2,-3,4}\right)$$ is applied to point $$B\left({3,-1,5}\right).$$ Determine the moment of this force relative to point $$A\left({1,2,-2}\right).$$

### Example 3

Calculate the area of a triangle with vertices $$A\left({1,-2,0}\right),$$ $$B\left({3,1,-1}\right)$$ and $$C\left({2,-1,3}\right).$$

### Example 4

Calculate the area of a parallelogram built on the vectors $$\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$$ and $$\mathbf{b} = \mathbf{j} - \mathbf{i}.$$

### Example 5

Vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ are not collinear. Find the values of $$\lambda$$ for which the vectors $$\lambda\mathbf{a} + \mathbf{b}$$ and $$2\mathbf{a} + \lambda\mathbf{b}$$ are collinear.

### Example 6

Prove the identity

${\Vert{\mathbf{a} \times \mathbf{b}}\Vert}^2 = \left| {\begin{array}{*{20}{c}} \left({\mathbf{a}\cdot\mathbf{a}}\right) & \left({\mathbf{a}\cdot\mathbf{b}}\right)\\ \left({\mathbf{a}\cdot\mathbf{b}}\right) & \left({\mathbf{b}\cdot\mathbf{b}}\right) \end{array}} \right|$

### Example 1.

Given vectors $$\mathbf{a}\left({1,2,3}\right)$$ and $$\mathbf{b}\left({2,4,7}\right).$$ Find the coordinates of the cross product $$\mathbf{a} \times \mathbf{b}.$$

Solution.

We determine the cross product by the formula

$\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}} \end{array}} \right|.$

Substituting the coordinates of vectors $$\mathbf{a}$$ and $$\mathbf{b},$$ we get:

$\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 2 & 3\\ 2 & 4 & 7 \end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}} 2 & 3\\ 4 & 7 \end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}} 1 & 3\\ 2 & 7 \end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}} 1 & 2\\ 2 & 4 \end{array}} \right| = \left({14-12}\right)\mathbf{i} - \left({7-6}\right)\mathbf{j} + \left({4-4}\right)\mathbf{k} = 2\mathbf{i} - \mathbf{j}.$

Therefore, the coordinates of the vector $$\mathbf{c} = \mathbf{a} \times \mathbf{b}$$ are $$\left({2,-1,0}\right).$$

### Example 2.

Force $$\mathbf{F}\left({2,-3,4}\right)$$ is applied to point $$B\left({3,-1,5}\right).$$ Determine the moment of this force relative to point $$A\left({1,2,-2}\right).$$

Solution.

The moment of force $$\mathbf{F}$$ about a point $$A$$ is defined as the cross product:

$\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F}.$

Calculate the position vector $$\mathbf{r_{AB}}:$$

$\mathbf{r_{AB}} = \underbrace{\left({3,-1,5}\right)}_{B} - \underbrace{\left({1,2,-2}\right)}_{A} = \left({2,-3,7}\right).$

The moment of the force is

$\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -3 & 7\\ 2 & -3 & 4 \end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}} -3 & 7\\ -3 & 4 \end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}} 2 & 7\\ 2 & 4 \end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}} 2 & -3\\ 2 & -3 \end{array}} \right| = \left({-12+21}\right)\mathbf{i} - \left({8-14}\right)\mathbf{j} + 0\cdot\mathbf{k} = 9\mathbf{i} + 6\mathbf{j}.$

Hence, $$\mathbf{M_A} = \left({9,6,0}\right).$$

### Example 3.

Calculate the area of a triangle with vertices $$A\left({1,-2,0}\right),$$ $$B\left({3,1,-1}\right)$$ and $$C\left({2,-1,3}\right).$$

Solution.

The area of the triangle is equal to half the area of the parallelogram spanned by the vectors AB and AC, that is

$A_{\triangle} = \frac{1}{2}\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert.$

Calculate the coordinates of vectors AB and AC:

$\mathbf{AB} = \underbrace{\left({3,1,-1}\right)}_B - \underbrace{\left({1,-2,0}\right)}_A = \left({2,3,-1}\right);$
$\mathbf{AC} = \underbrace{\left({2,-1,3}\right)}_C - \underbrace{\left({1,-2,0}\right)}_A = \left({1,1,3}\right);$

Find the cross product:

$\mathbf{AB} \times \mathbf{AC} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & -1\\ 1 & 1 & 3 \end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}} 3 & -1\\ 1 & 3 \end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}} 2 & -1\\ 1 & 3 \end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}} 2 & 3\\ 1 & 1 \end{array}} \right| = \left({9+1}\right)\mathbf{i} - \left({6+1}\right)\mathbf{j} + \left({2-3}\right)\mathbf{k} = 10\mathbf{i} - 7\mathbf{j} - \mathbf{k}.$

The absolute value of this vector is

$\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert = \sqrt{10^2 + \left({-7}\right)^2 + \left({-1}\right)^2} = \sqrt{150} =5\sqrt{6}.$

Then the area of the triangle is

$A_{\triangle} = \frac{1}{2}\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert = \frac{5\sqrt{6}}{2}.$

### Example 4.

Calculate the area of a parallelogram built on the vectors $$\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$$ and $$\mathbf{b} = \mathbf{j} - \mathbf{i}.$$

Solution.

Determine the cross product of vectors $$\mathbf{a}$$ and $$\mathbf{b}:$$

$\mathbf{a} \times \mathbf{b} = \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \left({\mathbf{j} - \mathbf{i}}\right) = \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \mathbf{j} - \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \mathbf{i}.$

Since cross product is anticommutative, we have

$\mathbf{a} \times \mathbf{b} = \mathbf{i} \times \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) - \mathbf{j} \times \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right).$

By the distributive property,

$\mathbf{a} \times \mathbf{b} = \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{j} + \mathbf{i} \times \mathbf{k} - \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{j} - \mathbf{j} \times \mathbf{k}.$

Substitute the results of multiplication of the unit vectors:

$\mathbf{a} \times \mathbf{b} = \mathbf{0} + \mathbf{k} + \left({-\mathbf{j}}\right) - \left({-\mathbf{k}}\right) - \mathbf{0} - \mathbf{i} = -\mathbf{i} - \mathbf{j} + 2\mathbf{k}.$

Now we can calculate the area of the parallelogram:

$A = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{-\mathbf{i} - \mathbf{j} + 2\mathbf{k}}\Vert = \sqrt{\left({-1}\right)^2 + \left({-1}\right)^2 + 2^2} = \sqrt{6}.$

### Example 5.

Vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ are not collinear. Find the values of $$\lambda$$ for which the vectors $$\lambda\mathbf{a} + \mathbf{b}$$ and $$2\mathbf{a} + \lambda\mathbf{b}$$ are collinear.

Solution.

If two vectors are collinear then their cross product (denote it by c) is equal to the zero vector. Hence

$\mathbf{c} = \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times \left({2\mathbf{a} + \lambda\mathbf{b}}\right) = \mathbf{0}.$

Using the distributive property, we can write:

$\mathbf{c} = \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times 2\mathbf{a} + \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times \lambda\mathbf{b} = \mathbf{0}.$

The cross product is anticommutative, so

$\mathbf{c} = - 2\mathbf{a} \times \left({\lambda\mathbf{a} + \mathbf{b}}\right) - \lambda\mathbf{b} \times \left({\lambda\mathbf{a} + \mathbf{b}}\right) = \mathbf{0}.$

Apply again the distributive property:

$\mathbf{c} = - \underbrace{2\mathbf{a} \times \lambda\mathbf{a}}_1 - \underbrace{2\mathbf{a} \times \mathbf{b}}_2 - \underbrace{\lambda\mathbf{b} \times \lambda\mathbf{a}}_3 - \underbrace{\lambda\mathbf{b} \times \mathbf{b}}_4 = \mathbf{0}.$

Consider the terms $$1$$ and $$4:$$

$2\mathbf{a} \times \lambda\mathbf{a} = 2\lambda\left({\mathbf{a} \times \mathbf{a}}\right) = 2\lambda \times \mathbf{0} = \mathbf{0};$
$\lambda\mathbf{b} \times \mathbf{b} = \lambda\left({\mathbf{b} \times \mathbf{b}}\right) = \lambda \times \mathbf{0} = \mathbf{0};$

Therefore

$\mathbf{c} = - 2\mathbf{a} \times \mathbf{b} - \lambda\mathbf{b} \times \lambda\mathbf{a} = \mathbf{0},$
$\Rightarrow \mathbf{c} = - 2\left({\mathbf{a} \times \mathbf{b}}\right) + \lambda\left({\mathbf{a} \times \mathbf{b}}\right) = \mathbf{0},$

or

$\left({\lambda - 2}\right)\left({\mathbf{a} \times \mathbf{b}}\right) = \mathbf{0}.$

From the description of the problem it is known that $${\mathbf{a} \times \mathbf{b} \ne \mathbf{0}}.$$ Therefore

$\lambda - 2 = 0, \Rightarrow \lambda = 0.$

### Example 6.

Prove the identity

${\Vert{\mathbf{a} \times \mathbf{b}}\Vert}^2 = \left| {\begin{array}{*{20}{c}} \left({\mathbf{a}\cdot\mathbf{a}}\right) & \left({\mathbf{a}\cdot\mathbf{b}}\right)\\ \left({\mathbf{a}\cdot\mathbf{b}}\right) & \left({\mathbf{b}\cdot\mathbf{b}}\right) \end{array}} \right|$

Solution.

Expand the determinant on the right side:

$RHS = \left| {\begin{array}{*{20}{c}} \left({\mathbf{a}\cdot\mathbf{a}}\right) & \left({\mathbf{a}\cdot\mathbf{b}}\right)\\ \left({\mathbf{a}\cdot\mathbf{b}}\right) & \left({\mathbf{b}\cdot\mathbf{b}}\right) \end{array}} \right| = \left({\mathbf{a}\cdot\mathbf{a}}\right)\left({\mathbf{b}\cdot\mathbf{b}}\right) - \left({\mathbf{a}\cdot\mathbf{b}}\right)\left({\mathbf{a}\cdot\mathbf{b}}\right).$

Remember that

$\mathbf{a}\cdot\mathbf{a} = {\Vert{\mathbf{a}}\Vert}^2,\;\mathbf{b}\cdot\mathbf{b} = {\Vert{\mathbf{b}}\Vert}^2,\;\mathbf{a}\cdot\mathbf{b} = {\Vert{\mathbf{a}}\Vert}{\Vert{\mathbf{b}}\Vert}\cos\varphi,\;$

where $$\varphi$$ is the angle between vectors $$\mathbf{a}$$ and $$\mathbf{b}.$$ Then

$RHS = {\Vert{\mathbf{a}}\Vert}^2 {\Vert{\mathbf{b}}\Vert}^2 - {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\cos^2\varphi = {\Vert{\mathbf{a}}\Vert}^2 {\Vert{\mathbf{b}}\Vert}^2 \left({1 - \cos^2\varphi}\right) = {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\sin^2\varphi$

It's clear that

$RHS = {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\sin^2\varphi = \left({{\Vert{\mathbf{a}}\Vert}{\Vert{\mathbf{b}}\Vert}\sin\varphi}\right)^2 = \left({\mathbf{a} \times \mathbf{b}}\right)^2 = {\Vert{\mathbf{a} \times \mathbf{b}}\Vert}^2 = LHS. \blacksquare$