In order to understand the material on this page, you need to know some linear algebra, specifically, how to calculate the determinant of 2x2 and 3x3 matrices. You can find a quick review on the linear algebra page.
Calculating The Cross Product 
The Cross Product is one way to 'multiply' two vectors (the other way is the dot product). Unlike the dot product, the cross product only makes sense when performed on two 3dim vectors. Taking the cross product of the two vectors \( 3\hat{i}+2\hat{j} \) and \( \hat{i}+\hat{j} \) is not possible, unless you mean \( 3\hat{i}+2\hat{j}+0\hat{k} \) and \( \hat{i}+\hat{j}+0\hat{k} \), in which case, you need to write out the \(\hat{k}\) term even if it is zero.
If we have two vectors, \(\vec{u}=\langle u_1, u_2, u_3 \rangle \) and \(\vec{v} = \langle v_1, v_2, v_3 \rangle \), we write the cross product of these two vectors as \( \vec{u} \times \vec{v} \).
The result of the cross product of two vectors is another vector. It's meaning is discussed later on this page. For now, let's focus on how we calculate the cross product.
To calculate the cross product, we use some linear algebra. If you haven't already, now would be good time to review the linear algebra page to make sure your skills calculating a 3x3 determinant are sharp. To calculate the cross product we calculate the following determinant.
\(\vec{u} \times \vec{v} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}\)
\( = (u_2 v_3  u_3 v_2)\hat{i}  (u_1 v_3  u_3 v_1)\hat{j} + (u_1 v_2  u_2 v_1)\hat{k} \)
Comments 
1. It is best NOT to memorize the last expression. Instead, set up and evaluate the determinant.
2. Remember to subtract the middle term.
3. It is important to set up the determinant correctly, i.e.
 The first row is the set of unit vectors.
 The second row is the first vector of the cross product.
 The third row is the second vector of the cross product.
The rows cannot be in any other order (more on this in the properties section below).
The name 'cross product' comes from the notation using '\(\times\)' between the two vectors. Just like with the dot product, it is important to use the '\(\times\)' between the vectors to indicate a cross product. Writing \(\vec{u} \vec{v} \) makes no sense and is considered incorrect notation.
Note: Recently we heard that what we call the 'determinant' above is not strictly a determinant but just a mnemonic device to calculate the cross product, since a true determinant consists of only numbers not vectors. We have not verified this at this time but it certainly could be true. When we verify it, we will update this page accordingly.
Okay, so let's watch a video clip discussing the cross product and its geometric interpretation.
video by PatrickJMT 

Problem Statement 

Given \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \), calculate the cross product \( \vec{A} \times \vec{B} \).
Final Answer 

Problem Statement 

Given \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \), calculate the cross product \( \vec{A} \times \vec{B} \).
Solution 

Setting up the determinant, we have
\(\vec{A} \times \vec{B} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & 0 & 3
\end{vmatrix} = \) \((90)\hat{i}  (64)\hat{j} + (03)\hat{k} = \) \( 9\hat{i}  2\hat{j}  3\hat{k} \)
Final Answer 

\( (2\hat{i} + 3\hat{j} + 4\hat{k} ) \times ( \hat{i} + 3\hat{k} ) = 9\hat{i}  2\hat{j}  3\hat{k} \) 
close solution

Problem Statement 

Given \( \vec{A} = \hat{i}2\hat{j}+2\hat{k} \) and \( \vec{B} = 3\hat{i}\hat{j}\hat{k} \), calculate the cross product \(\vec{A}\times\vec{B}\).
Final Answer 

Problem Statement 

Given \( \vec{A} = \hat{i}2\hat{j}+2\hat{k} \) and \( \vec{B} = 3\hat{i}\hat{j}\hat{k} \), calculate the cross product \(\vec{A}\times\vec{B}\).
Solution 

\(\vec{A} \times \vec{B} = (\hat{i}2\hat{j}+2\hat{k}) \times (3\hat{i}\hat{j}\hat{k}) = \) \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \) \( (2 (2))\hat{i}  (16)\hat{j} + (1(6))\hat{k} = \) \( 4\hat{i} + 7\hat{j} +5\hat{k} \)
Final Answer 

\( \vec{A} \times \vec{B} = 4\hat{i} + 7\hat{j} + 5\hat{k} \) 
close solution

Problem Statement 

Calculate the cross product of \(\hat{i}\) and \(\hat{j}\).
Final Answer 

Problem Statement 

Calculate the cross product of \(\hat{i}\) and \(\hat{j}\).
Solution 

Before doing the calculation, stop and think for a minute what the answer might be. Think about the properties of the cross product and the right hand rule. Do you have an idea of what the answer might be?
Okay, let's do the calculation.
\( \hat{i} \times \hat{j} = \) \(
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 0 \\
0 & 1 & 0
\end{vmatrix} = \)
\( (00)\hat{i}  (00)\hat{j} + (10)\hat{k} = \hat{k}\)
Let's think about this answer for a minute. Notice that the standard unit vectors are vectors of length one that coincide with the Cartesian coordinate system. All of the vectors are perpendicular to one another, so it makes sense that the vector perpendicular to \(\hat{i}\) and \(\hat{j}\) would be in the direction of the zaxis or \(\hat{k}\).
A question to consider: If you take the cross product of two unit vectors, will the result always be a unit vector? Why or why not?
Final Answer 

\(\hat{i}\times\hat{j}=\hat{k}\) 
close solution

Problem Statement 

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

video by Krista King Math 

close solution

Problem Statement 

Calculate the cross product of the vectors \( \vec{a} = \hat{i}  \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k} \).
Solution 

video by Krista King Math 

close solution

Problem Statement 

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

video by Krista King Math 

close solution

Problem Statement 

Calculate the cross product of the vectors \( \vec{a} = \langle 5,1,4 \rangle \) and \( \vec{b} = \langle 1,0,2 \rangle \).
Solution 

video by PatrickJMT 

close solution

Cross Product Properties 
Here are some cross product properties.
Algebraic Properties
Let \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\) be vectors in space and let \(a\) be a scalar.
1. \(\vec{u} \times \vec{v} = (\vec{v} \times \vec{u})\)
2. \(\vec{u} \times (\vec{v}+\vec{w}) = (\vec{u} \times \vec{v}) + (\vec{u} \times \vec{w}) \)
3. \(a(\vec{u} \times \vec{v}) = (a\vec{u}) \times \vec{v} = \vec{u} \times (a\vec{v})\)
4. \(\vec{u} \times \vec{0} = \vec{0} \)
5. \( \vec{u} \times \vec{u} = \vec{0} \)
Notes:
 property 1 implies that the cross product is not commutative
 in property 2, the vector \(\vec{u}\) is on the left, so when it is distributed across the addition, it must remain on the left in both cases
 property 5 seems trivial but it is very powerful; later on this page, this will be discussed in more detail.
Here is a video with proofs of some of these algebraic properties.
video by Larson Calculus 

Geometric Properties
Let \(\vec{u}\) and \(\vec{v}\) be nonzero vectors in space and let \(\theta\) be the angle between \(\vec{u}\) and \(\vec{v}\).
6. \(\vec{u} \times \vec{v}\) is orthogonal to both \(\vec{u}\) and \(\vec{v}\).
7. \( \ \vec{u} \times \vec{v} \ = \\vec{u}\ \\vec{v}\ \sin \theta\)
8. \(\vec{u} \times \vec{v} = \vec{0} \) if and only if \(\vec{u}\) is a scalar multiple of \(\vec{v}\)
9. \( \ \vec{u} \times \vec{v} \ \) represents the area of the parallelogram formed with \(\vec{u}\) and \(\vec{v}\) as adjacent sides.
Notes:
 for the geometric properties, both vectors must be nonzero; this is not a requirement in the algebraic properties
 notice in property 7, the cross product involves the sine of angle \(\theta\) while the dot product involves the cosine of the angle
 as mentioned in property 9, the cross product is the area of a parallelogram; here is a great video that discusses this in more detail.
video by Dr Chris Tisdell 

Here is a video with proofs of some of these geometric properties.
video by Larson Calculus 

There is a simple rule to use when you need to know the direction of the resulting vector from the cross product. It's called the right hand rule. The idea is to lay out your hand with all fingers straight out. Place the middle of your hand at the point of intersection of the two vectors involved in the cross product with your fingers in the direction of the first vector. Fold your fingers in the direction of the second vector. Your thumb will then be pointing in the direction of the result of the cross product. Here is a quick video showing this idea.
video by Animations for Physics and Astronomy 

All this information may be a bit overwhelming. So let's take a few minutes and watch this video. He explains the cross product very well and shows some examples.
video by Dr Chris Tisdell 

Okay, let's use some of these properties to solve these problems.
Problem Statement 

Given \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \), use the cross product to find the angle between \(\vec{A}\) and \(\vec{B}\).
Final Answer 

Problem Statement 

Given \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{B} = \hat{i} + 3\hat{k} \), use the cross product to find the angle between \(\vec{A}\) and \(\vec{B}\).
Solution 

From a previous problem, we found the cross product as \(\vec{A} \times \vec{B} = 9\hat{i}  2\hat{j}  3\hat{k} \)
To find the angle between the two vectors, we use the formula \( \ \vec{A} \times \vec{B} \ = \\vec{A}\ \\vec{B}\ \sin (\theta)\), where \(\theta\) is the angle between the two vectors.
Let's calculate these magnitudes.
\( \ \vec{A} \times \vec{B} \ \) 
\( \ 9\hat{i}  2\hat{j}  3\hat{k} \ \) 
\( \sqrt{9^2 + (2)^2 + (3)^2} \) 
\( \sqrt{81+4+9} = \sqrt{94} \) 
\( \\vec{A} \ = \sqrt{2^2+3^2+4^2} = \sqrt{4+9+16} = \sqrt{29}\)
\( \\vec{B}\ = \sqrt{1^2+0^2+3^2} = \sqrt{10}\)
\(\displaystyle{ \sin(\theta) = \frac{\ \vec{A} \times \vec{B} \}{\\vec{A}\ \\vec{B}\} }\) 
\(\displaystyle{ \sin(\theta) = \frac{\sqrt{94}}{\sqrt{29}\sqrt{10}} }\) 
\(\displaystyle{ \theta = \arcsin\left( \frac{\sqrt{94}}{\sqrt{290}} \right) \approx 0.606 }\) 
Final Answer 

The angle between the vectors \( 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \hat{i} + 3\hat{k} \) is approximately 0.606 radians. 
close solution

Problem Statement 

Find a unit vector that is perpendicular to \( \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{B} = 3\hat{i} + 2\hat{j} + \hat{k} \).
Final Answer 

Problem Statement 

Find a unit vector that is perpendicular to \( \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{B} = 3\hat{i} + 2\hat{j} + \hat{k} \).
Solution 

Note: There are two possible answers to this question.
Since the result of the cross product is perpendicular to both of the vectors used to calculate the cross product, we can calculate the cross product and then find the unit vector of the result.
So, let's start by finding the cross product \(\vec{A}\times\vec{B}\).
\(\vec{A}\times\vec{B} = \) \( (\hat{i}+2\hat{j}+3\hat{k}) \times (3\hat{i}+2\hat{j}+\hat{k}) = \) \(
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
3 & 2 & 1
\end{vmatrix} = \) \( (26)\hat{i}  (19)\hat{j} + (26)\hat{k} = \) \( 4\hat{i} + 8\hat{j}  4\hat{k} \)
Okay, now we need to find the unit vector in the same direction as the result of the cross product. Let's call this vector \(\vec{v} =4\hat{i} + 8\hat{j}  4\hat{k} \).
\(\displaystyle{ \vec{u} = \frac{1}{\\vec{v}\}\vec{v} }\) 
\(\displaystyle{ \frac{1}{\sqrt{(4)^2+8^2+(4)^2}}(4\hat{i} + 8\hat{j}  4\hat{k}) }\) 
\(\displaystyle{ \frac{1}{\sqrt{16+64+16}}(4\hat{i} + 8\hat{j}  4\hat{k}) }\) 
\(\displaystyle{ \frac{1}{\sqrt{96}}(4\hat{i} + 8\hat{j}  4\hat{k}) }\) 
\(\displaystyle{ \frac{1}{4\sqrt{6}}(4\hat{i} + 8\hat{j}  4\hat{k}) }\) 
\(\displaystyle{ \frac{1}{\sqrt{6}}(\hat{i} + 2\hat{j}  \hat{k}) }\) 
Note: If you calculated \(\vec{B}\times\vec{A}\), your unit vector would be \(\vec{u}\), which is the other correct answer.
Final Answer 

\(\displaystyle{ \vec{u} = \frac{\pm 1}{\sqrt{6}} (\hat{i}+2\hat{j}\hat{k}) }\) 
close solution

Problem Statement 

Show that the vectors \(\vec{A} = 2\hat{i}3\hat{j}+4\hat{k}\), \(\vec{B} = 6\hat{i}+2\hat{j}+\hat{k}\) and \(\vec{C} = 6\hat{i}+10\hat{j}7\hat{k}\) are coplanar.
Final Answer 

Problem Statement 

Show that the vectors \(\vec{A} = 2\hat{i}3\hat{j}+4\hat{k}\), \(\vec{B} = 6\hat{i}+2\hat{j}+\hat{k}\) and \(\vec{C} = 6\hat{i}+10\hat{j}7\hat{k}\) are coplanar.
Solution 

We know that if the 3 vectors are coplanar, then the triple scalar product is zero. Although not explicitly stated here, we can assume the reverse, i.e. if the triple scalar product is zero, then the vectors are coplanar. [Note: The assumption of the reverse being true cannot be always be made. Be careful when making this assumption.]
So let's calculate the triple scalar product to see if it is zero.
\( \vec{A}\cdot \vec{B} \times \vec{C} = 2\hat{i}3\hat{j}+4\hat{k} \cdot 6\hat{i}+2\hat{j}+\hat{k} \times 6\hat{i}+10\hat{j}7\hat{k} = \) \(\begin{vmatrix}
2 & 3 & 4 \\
6 & 2 & 1 \\
6 & 10 & 7
\end{vmatrix} = \) \( 2(1410)  (3)(42  6) + 4(6012) = 0 \)
Final Answer 

Since the triple scalar product is zero, the vectors are coplanar. 
close solution

Problem Statement 

Show that the vectors \( \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{B} = 6\hat{i} + 2\hat{j}  3\hat{k} \) are perpendicular.
Final Answer 

Problem Statement 

Show that the vectors \( \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{B} = 6\hat{i} + 2\hat{j}  3\hat{k} \) are perpendicular.
Solution 

Perpendicular vectors have a angle of \(\pi/2\) between them. So, we will calculate the angle between the vectors and determine if it is \(\pi/2\). Actually, since we know that \(\\vec{A} \times \vec{B}\ = \\vec{A}\ \\vec{B}\ \sin(\theta)\) and when \(\theta = \pi/2\), \(\sin(\theta) = 1\), we can calculate \(\\vec{A} \times \vec{B}\\) and see if it is equal to \(\\vec{A}\ \\vec{B}\\). If they are equal, then the vectors are perpendicular.
\(\vec{A} \times \vec{B} = 2\hat{i}+3\hat{j}+6\hat{k} \times 6\hat{i} + 2 \hat{j}3\hat{k} = \)
\(\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
6 & 2 & 3
\end{vmatrix} = \) \( (9 12)\hat{i}  (6  36)\hat{j} + (4  18)\hat{k} = \) \( 21\hat{i} +42\hat{j} 14\hat{k} \)
\( \ \vec{A} \times \vec{B} \ = \) \(
\ 21\hat{i} +42\hat{j} 14\hat{k} \ = \) \(
\sqrt{(21)^2 + 42^2 + (14)^2} = \) \(
\sqrt{7^2 (9) + 7^2(36) + 7^2(4)} = 7\sqrt{49} = 49 \)
So \( \ \vec{A} \times \vec{B} \ = 49 \).
Now let's calculate \( \\vec{A}\ \\vec{B}\ \).
\( \\vec{A}\ \\vec{B}\ \) 
\( \ 2\hat{i}+3\hat{j}+6\hat{k} \ \ 6\hat{i} + 2 \hat{j}3\hat{k} \ \) 
\( \sqrt{2^2 + 3^2 + 6^2} \sqrt{6^2 + 2^2 + (3)^2} \) 
\( \sqrt{4 + 9 + 36} \sqrt{36 +4 + 9} \) 
\( \sqrt{49}\sqrt{49} = 7(7) = 49 \) 
Final Answer 

Since \( \\vec{A}\times\vec{B}\ = \\vec{A}\\\vec{B}\ \), the angle between the two vectors is \(\pi/2\) and so the vectors are perpendicular. 
close solution

Problem Statement 

Find a unit vector that is perpendicular to \( \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{B} = 6\hat{i} + 2\hat{j}  3\hat{k} \).
Final Answer 

Problem Statement 

Find a unit vector that is perpendicular to \( \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{B} = 6\hat{i} + 2\hat{j}  3\hat{k} \).
Solution 

Note: There are two possible answers to this problem.
Since the result of the cross product of two vectors is a vector perpendicular to the original two vectors, we can use the cross product to get a vector. Then, we just divide by the magnitude of the vector to get a unit vector in the same direction.
First, let's take the cross product.
\(\vec{A} \times \vec{B} = (2\hat{i}+3\hat{j}+6\hat{k}) \times (6\hat{i} + 2 \hat{j} 3\hat{k}) = \)
\(
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
6 & 2 & 3
\end{vmatrix} = \) \(
(9  12)\hat{i}  (6  36)\hat{j} + (4  18)\hat{k} = \) \(
21\hat{i} + 42\hat{j}  14\hat{k} = \) \(
7[ 3\hat{i}  6\hat{j} + 2\hat{k}]\)
We factored out the 7 before finding the magnitude to make the calculations easier. However, this is not necessary.
Now we need to calculate the unit vector in the same direction as this last vector.
\(\displaystyle{ \vec{u} = \frac{(7[ 3\hat{i}  6\hat{j} + 2\hat{k}])}{\ 7[ 3\hat{i}  6\hat{j} + 2\hat{k}] \ } }\) 
\(\displaystyle{ \frac{7[ 3\hat{i}  6\hat{j} + 2\hat{k}]}{\sqrt{(7)^2[3^2+(6)^2 +2^2]}} }\) 
\(\displaystyle{ \frac{7}{7\sqrt{9+36+4}}[ 3\hat{i}  6\hat{j} + 2\hat{k}] }\) 
\(\displaystyle{ \frac{1}{\sqrt{49}}[ 3\hat{i}  6\hat{j} + 2\hat{k}] }\) 
\(\displaystyle{ \frac{1}{7}[ 3\hat{i}  6\hat{j} + 2\hat{k}] }\) 
Notes: When solving this problem, we calculated \(\vec{A}\times\vec{B}\). If you calculated \(\vec{B}\times\vec{A}\), your answer would have been \(\vec{u}\), which is also a correct answer.
Final Answer 

The unit vectors perpendicular to \(2\hat{i}+3\hat{j}+6\hat{k}\) and \(6\hat{i}+2\hat{j}3\hat{k}\) are \(\displaystyle{\frac{\pm 1}{7}[3\hat{i}6\hat{j}+2\hat{k}]}\). 
close solution

Cross Product Applications 
Here is a list of some applications using the cross product. Some of them also require the use of the dot product.
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