# Cross Product of Two Vectors - Multiplying Vectors

## (Vector Product of Two Vectors)

The cross product, also called vector product of two vectors is written $$\vec{u}\times \vec{v}$$ and is the second way to multiply two vectors together.

When we multiply two vectors using the cross product we obtain a new vector. This is unlike the scalar product (or dot product) of two vectors, for which the outcome is a scalar (a number, not a vector!).

In fact, the cross product of two vectors $$\vec{u}$$ and $$\vec{v}$$ is a "new" vector that is perpendicular to both $$\vec{u}$$ and $$\vec{v}$$, we say that $$\vec{u}\times \vec{v}$$ is normal to the plane containing $$\vec{u}$$ and $$\vec{v}$$.

## How to Calculate the Cross Product

To calculate the vector product, or cross product, of two vectors we use either one of the following two options:

• Option 1: use the Formula (learn it off by heart)
• Option 2: use Matrix Algebra (recommended method)
We look at both options here.

### Option 1 - The Formula: Vector Product $$\vec{u}\times \vec{v}$$

Given two vectors $$\vec{u} = \begin{pmatrix}u_1 \\ u_2 \\ u_3 \end{pmatrix}$$ and $$\vec{v} = \begin{pmatrix}v_1 \\ v_2 \\ v_3 \end{pmatrix}$$, the vector product, or cross product, $$\vec{u}\times \vec{v}$$ can be calculated using the following formula: $\vec{u} \times \vec{v} = \begin{pmatrix}u_2v_3 - v_2u_3\end{pmatrix} \vec{i} - \begin{pmatrix}u_1v_3 -v_1u_3\end{pmatrix} \vec{j} + \begin{pmatrix}u_1v_2 - v_1u_2\end{pmatrix} \vec{k}$ To avoid having the subtraction between the first and second term, some math courses (such as IB HL Mathematics) rewrite this formula as: $\vec{u} \times \vec{v} = \begin{pmatrix}u_2v_3 - v_2u_3\end{pmatrix} \vec{i} + \begin{pmatrix} v_1u_3 - u_1v_3\end{pmatrix} \vec{j} + \begin{pmatrix}u_1v_2 - v_1u_2\end{pmatrix} \vec{k}$ Both formula are completely equivalent.

## Example

Using the formula, stated above, find the cross product, $$\vec{u}\times \vec{v}$$ of $$\vec{u} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$$ and $$\vec{v} = \begin{pmatrix} 4 \\ 0 \\ 6 \end{pmatrix}$$.

#### Solution

Using the second formula, stated above: $\vec{u} \times \vec{v} = \begin{pmatrix}u_2v_3 - v_2u_3\end{pmatrix} \vec{i} + \begin{pmatrix} v_1u_3 - u_1v_3\end{pmatrix} \vec{j} + \begin{pmatrix}u_1v_2 - v_1u_2\end{pmatrix} \vec{k}$ With $$u_1 = 1$$, $$u_2 = -3$$, $$u_3 = 2$$, $$v_1 = 4$$, $$v_2 = 0$$ and $$v_3 = 6$$, we find: \begin{aligned} \vec{u}\times \vec{v} & = \begin{pmatrix}-3 \times 6 - 0 \times 2 \end{pmatrix}\vec{i} + \begin{pmatrix}4 \times 2 - 1 \times 6 \end{pmatrix}\vec{j} + \begin{pmatrix}1 \times 0 - 4 \times (-3)\end{pmatrix} \vec{k} \\ & = \begin{pmatrix}-18 -0\end{pmatrix}\vec{i} + \begin{pmatrix}8 - 6 \end{pmatrix} \vec{j} + \begin{pmatrix}0+12\end{pmatrix}\vec{k} \\ \vec{u}\times \vec{v} & = -18 \vec{i} + 2 \vec{j} + 12 \vec{k} \end{aligned}

### Option 2: Matrix Algebra - Determinant of a 3 by 3 Matrix (recommended)

Given two vectors $$\vec{u} = \begin{pmatrix}u_1 \\ u_2 \\ u_3 \end{pmatrix}$$ and $$\vec{v} = \begin{pmatrix}v_1 \\ v_2 \\ v_3 \end{pmatrix}$$, the vector product, or cross product, $$\vec{u}\times \vec{v}$$ equals to the following: $\vec{u}\times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$ Notice that:

• the components in the first row consist of the unit base vectors $$\vec{i}$$, $$\vec{j}$$ and $$\vec{k}$$ (this will always be the first row),
• the second row is the first vector in the product $$\vec{u}\times \vec{v}$$, so in this case $$\vec{u}$$,
• the third row is the second vector in the product $$\vec{u}\times \vec{v}$$, so in this case $$\vec{v}$$.

### Tutorial: Cross Product - Determinant Method

Given the two vectors $$\vec{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$$ and $$\vec{b} = \begin{pmatrix} 5 \\ 0 \\ 4 \end{pmatrix}$$, we learn how to calculate the cross product, $$\vec{a}\times \vec{b}$$, using Matrix Algebra.

## Example

Given $$\vec{u} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$$ and $$\vec{v} = \begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix}$$, calculate their cross product $$\vec{u}\times \vec{v}$$.

### Solution

Using the method we just saw, we can state: \begin{aligned} \vec{u} \times \vec{v} & = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -3 \\ 4 & 0 & 5 \end{vmatrix} \\ & = \vec{i} \begin{vmatrix} 1 & -3 \\ 0 & 5 \end{vmatrix} - \vec{j} \begin{vmatrix} 2 & -3 \\ 4 & 5\end{vmatrix} + \vec{k} \begin{vmatrix} 2 & 1 \\ 4 & 0 \end{vmatrix} \\ & = \vec{i} \begin{pmatrix} 1 \times 5 - 0 \times (-3)\end{pmatrix} - \vec{j} \begin{pmatrix} 2\times 5 - 4 \times (-3) \end{pmatrix} + \vec{k} \begin{pmatrix} 2\times 0 - 4 \times 1 \end{pmatrix} \\ & = \vec{i}\begin{pmatrix} 5 - 0\end{pmatrix} - \vec{j} \begin{pmatrix} 10 - (-12)\end{pmatrix} + \vec{k} \begin{pmatrix} 0 - 4 \end{pmatrix} \\ & = \vec{i} \begin{pmatrix} 5 \end{pmatrix} - \vec{j}\begin{pmatrix} 10+12 \end{pmatrix} + \vec{k} \begin{pmatrix} - 4 \end{pmatrix} \\ \vec{u} \times \vec{v} & = 5 \vec{i} - 22 \vec{j} - 4 \vec{k} \end{aligned} That's our final answer, we can now state that $$\vec{u} \times \vec{v} = 5 \vec{i} - 22 \vec{j} - 4 \vec{k}$$.

## Exercise 1

1. Find $$\vec{a} \times \vec{b}$$, given $$\vec{a} = \begin{pmatrix}2 \\ 3 \\ 4 \end{pmatrix}$$ and $$\vec{b} = \begin{pmatrix} 5 \\ -2 \\ 1 \end{pmatrix}$$.

2. Find $$\vec{c}\times \vec{d}$$, given $$\vec{c} = 3 \vec{i} + 5 \vec{j} - \vec{k}$$ and $$\vec{d} = \vec{i} -4 \vec{j} +2 \vec{k}$$.

3. Find $$\vec{u}\times \vec{v}$$, given $$\vec{u} = 6 \vec{i} - 3 \vec{k}$$ and $$\vec{v} = -2 \vec{i} + \vec{j} + 5 \vec{k}$$.

4. Find $$\vec{a} \times \vec{b}$$, given $$\vec{a} = \begin{pmatrix}-2 \\ 0 \\ 6 \end{pmatrix}$$ and $$\vec{b} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$$.

5. Find $$\vec{m}\times \vec{n}$$, given $$\vec{m} = -3\vec{i} + 5 \vec{j} - \vec{k}$$ and $$\vec{n} = 4\vec{j} + 7 \vec{k}$$.

6. Find $$\vec{u}\times \vec{v}$$, given $$\vec{u} = \begin{pmatrix}8 \\ - 1 \\ -2 \end{pmatrix}$$ and $$\vec{v} = \begin{pmatrix}-3 \\ - 2 \\ -4 \end{pmatrix}$$.

7. Find $$\vec{a}\times \vec{b}$$, given $$\vec{a} = 3 \vec{i} - 2 \vec{j}$$ and $$\vec{b} = 5 \vec{i} + \vec{j}$$.

8. Find $$\vec{c}\times \vec{d}$$, given $$\vec{c} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$$ and $$\vec{d} = \begin{pmatrix} 0 \\ 3 \\ 5 \end{pmatrix}$$.

Note: this exercise can be downloaded as a worksheet to practice with:
The Rest of the Page is under construction and will be finished very very very soon :)

## Exercise 2 (Exam Style Questions)

1. The vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ satisfy the equation $$\vec{a}+\vec{b} + \vec{c} = \vec{0}$$. Show that $$\vec{a}\times \vec{b} = \vec{b}\times \vec{c} = \vec{c}\times \vec{a}$$.