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17calculus > vectors > cross product

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 3-dim 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.

$$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$$ $$= (u_2 v_3 - u_3 v_2)\hat{i} - (u_1 v_3 - u_3 v_1)\hat{j} + (u_1 v_2 - u_2 v_1)\hat{k}$$

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.

PatrickJMT - geometric interpretation of the cross product [2mins-25secs]

video by PatrickJMT

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

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 $$\vec{A} \times \vec{B}$$.

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

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
$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 0 & 3 \end{vmatrix} =$$ $$(9-0)\hat{i} - (6-4)\hat{j} + (0-3)\hat{k} =$$ $$9\hat{i} - 2\hat{j} - 3\hat{k}$$

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

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 $$\vec{A}\times\vec{B}$$.

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 $$\vec{A}\times\vec{B}$$.

$$\vec{A} \times \vec{B} = 4\hat{i} + 7\hat{j} + 5\hat{k}$$

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

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

$$\vec{A} \times \vec{B} = 4\hat{i} + 7\hat{j} + 5\hat{k}$$

Calculate the cross product of $$\hat{i}$$ and $$\hat{j}$$.

Problem Statement

Calculate the cross product of $$\hat{i}$$ and $$\hat{j}$$.

$$\hat{i}\times\hat{j}=\hat{k}$$

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.
$$\hat{i} \times \hat{j} =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} =$$ $$(0-0)\hat{i} - (0-0)\hat{j} + (1-0)\hat{k} = \hat{k}$$
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 z-axis 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?

$$\hat{i}\times\hat{j}=\hat{k}$$

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

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

41 solution video

video by Krista King Math

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

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

42 solution video

video by Krista King Math

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

Problem Statement

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

Solution

43 solution video

video by Krista King Math

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

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

45 solution video

video by PatrickJMT

 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.

Larson Calculus - proofs of some algebraic properties [2mins-3secs]

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.

Dr Chris Tisdell - Cross Product and Area of Parallelogram [9mins]

video by Dr Chris Tisdell

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

Larson Calculus - proofs [3mins-39secs]

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.

Right Hand Rule for Cross Products [10secs]

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.

Dr Chris Tisdell - Cross product of vectors [20mins-24secs]

video by Dr Chris Tisdell

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

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 $$\vec{B}$$.

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 $$\vec{B}$$.

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.

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.

 $$\| \vec{A} \times \vec{B} \|$$ $$\| 9\hat{i} - 2\hat{j} - 3\hat{k} \|$$ $$\sqrt{9^2 + (-2)^2 + (-3)^2}$$ $$\sqrt{81+4+9} = \sqrt{94}$$

$$\|\vec{A} \| = \sqrt{2^2+3^2+4^2} = \sqrt{4+9+16} = \sqrt{29}$$
$$\|\vec{B}\| = \sqrt{1^2+0^2+3^2} = \sqrt{10}$$

 $$\displaystyle{ \sin(\theta) = \frac{\| \vec{A} \times \vec{B} \|}{\|\vec{A}\| \|\vec{B}\|} }$$ $$\displaystyle{ \sin(\theta) = \frac{\sqrt{94}}{\sqrt{29}\sqrt{10}} }$$ $$\displaystyle{ \theta = \arcsin\left( \frac{\sqrt{94}}{\sqrt{290}} \right) \approx 0.606 }$$

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.

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

Problem Statement

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

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

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 $$\vec{A}\times\vec{B}$$.
$$\vec{A}\times\vec{B} =$$ $$(\hat{i}+2\hat{j}+3\hat{k}) \times (3\hat{i}+2\hat{j}+\hat{k}) =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{vmatrix} =$$ $$(2-6)\hat{i} - (1-9)\hat{j} + (2-6)\hat{k} =$$ $$-4\hat{i} + 8\hat{j} - 4\hat{k}$$
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.

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

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.

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.

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

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.
$$\vec{A}\cdot \vec{B} \times \vec{C} = 2\hat{i}-3\hat{j}+4\hat{k} \cdot 6\hat{i}+2\hat{j}+\hat{k} \times 6\hat{i}+10\hat{j}-7\hat{k} =$$ $$\begin{vmatrix} 2 & -3 & 4 \\ 6 & 2 & 1 \\ 6 & 10 & -7 \end{vmatrix} =$$ $$2(-14-10) - (-3)(-42 - 6) + 4(60-12) = 0$$

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

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.

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.

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.

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 $$\|\vec{A}\| \|\vec{B}\|$$.

 $$\|\vec{A}\| \|\vec{B}\|$$ $$\| 2\hat{i}+3\hat{j}+6\hat{k} \| \| 6\hat{i} + 2 \hat{j}-3\hat{k} \|$$ $$\sqrt{2^2 + 3^2 + 6^2} \sqrt{6^2 + 2^2 + (-3)^2}$$ $$\sqrt{4 + 9 + 36} \sqrt{36 +4 + 9}$$ $$\sqrt{49}\sqrt{49} = 7(7) = 49$$

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.

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

Problem Statement

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

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 $$\displaystyle{\frac{\pm 1}{7}[3\hat{i}-6\hat{j}+2\hat{k}]}$$.

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.

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 $$\displaystyle{\frac{\pm 1}{7}[3\hat{i}-6\hat{j}+2\hat{k}]}$$.

 Cross Product Applications

Here are some applications using the cross product. Some of them also require the use of the dot product.

 Triple Scalar Product

The triple scalar product is a result of combining the dot product with the cross product. First, let's define what it is and then discuss a couple of properties.

Definition and Notation - If we have three vectors in space, $$\vec{u} = u_x\hat{i}+u_y\hat{j}+u_z\hat{k}$$, $$\vec{v} = v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ and $$\vec{w} = w_x\hat{i}+w_y\hat{j}+w_z\hat{k}$$, then the triple scalar product is defined to be $$\vec{u} \cdot (\vec{v} \times \vec{w})$$
The calculation of this can be done as follows
$$\vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$$

Triple Scalar Product Proof

Theorem: Triple Scalar Product

If we have three vectors in space,
$$\vec{u} = u_x\hat{i}+u_y\hat{j}+u_z\hat{k}$$, $$\vec{v} = v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$$ and $$\vec{w} = w_x\hat{i}+w_y\hat{j}+w_z\hat{k}$$,
then the triple scalar product is $$\vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$$

Proof - - To prove this, we will calculate $$\vec{u} \cdot (\vec{v} \times \vec{w})$$ and then calculate the determinant to show that we get the same result. Let's start with the cross product.

$$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} =$$ $$(v_y w_z - v_z w_y)\hat{i} -$$ $$(v_x w_z - w_x v_z)\hat{j} +$$ $$(v_x w_y - w_x v_y)\hat{k}$$

Taking the dot product of $$\vec{u}$$ and the last equation gives us

$$(u_x\hat{i}+u_y\hat{j}+u_z\hat{k}) \cdot [ (v_y w_z - v_z w_y)\hat{i} -$$ $$(v_x w_z - w_x v_z)\hat{j} +$$ $$(v_x w_y - w_x v_y)\hat{k} ] =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y)$$

We could certainly multiply the components of vector $$\vec{u}$$ through each factor, but for reasons you will see later, we will leave the equation as it is. To sum up, we have calculated

$$\vec{u} \cdot (\vec{v} \times \vec{w}) =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y) ~~~~~ (1)$$

Okay, now let's calculate the determinant
$$\begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$$

We will go across the top row (although going down the first column will give us the same result, going across the top row makes the algebra come out the way want it to).

$$\begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix} =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y)$$

Notice this last equation is the same as equation (1) above. This is easier to see since we did not multiply out all factors in Equation (1). So, we have shown

$$\vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$$       [qed]

The triple scalar product is so named because the result is a scalar. [For comparison, see the triple vector product section below.]

Properties - The triple scalar product can also be evaluated in one of the following forms.
$$\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})$$
The parentheses may be omitted since evaluating the dot product first yields a scalar and it doesn't make sense to take the cross product of a scalar with a vector.
This property also holds $$[\vec{a} \cdot (\vec{b} \times \vec{c})]\vec{a} = (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})$$
When the triple scalar product is zero, the 3 vectors are coplanar.

Applications - Geometrically, the triple scalar product $$\vec{a} \cdot (\vec{b} \times \vec{c} )$$ is the (signed) volume of the parallelepiped defined by the three vectors given (see figure on the right). The word 'signed' means that the result can be positive or negative depending on the orientation of the vectors.
You can probably now see that when the 3 vectors are coplanar, the parallelepiped is flat and has no volume, so the triple scalar product is zero.

Other Names For the Triple Scalar Product -
- scalar triple product
- mixed product
- box product

Here are several videos that explains this in more detail. The first two are especially good and the third contains a proof.

Dr Chris Tisdell - Scalar triple product [9mins-30secs]

video by Dr Chris Tisdell

Dr Chris Tisdell - Scalar triple product and volume [11mins-49secs]

video by Dr Chris Tisdell

Larson Calculus - Proof - Geometric Property of the Triple Scalar Product [1mins-40secs]

video by Larson Calculus

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B}=3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple scalar product $$\vec{A}\cdot\vec{B}\times\vec{C}$$.

Problem Statement

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B}=3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple scalar product $$\vec{A}\cdot\vec{B}\times\vec{C}$$.

$$\vec{A}\cdot\vec{B}\times\vec{C}=-9$$

Problem Statement

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B}=3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple scalar product $$\vec{A}\cdot\vec{B}\times\vec{C}$$.

Solution

First, we need to find the cross product.

$$\vec{B}\times\vec{C}$$ = $$(3\hat{i}-\hat{j}-\hat{k}) \times (-\hat{i}-\hat{k}) =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & -1 \\ -1 & 0 & -1 \end{vmatrix}$$ $$= (1-0)\hat{i} - (-3-1)\hat{j} + (0-1)\hat{k}$$ $$= \hat{i} +4\hat{j} -\hat{k}$$

Now we do the dot product.
$$(\hat{i}-2\hat{j}+2\hat{k}) \cdot (\hat{i} +4\hat{j} -\hat{k}) =$$ $$(1)(1) + (-2)(4) + (2)(-1) =$$ $$1-8-2 = -9$$

$$\vec{A}\cdot\vec{B}\times\vec{C}=-9$$

 Triple Vector Product

The triple vector product (or vector triple product, as it is sometimes called) is so named because the result is a vector. [For comparison, see the triple scalar product panel above.]

When you have three vectors, $$\vec{u}$$, $$\vec{v}$$ and $$\vec{w}$$, the triple vector product is defined as $$\vec{u} \times \vec{v} \times \vec{w}$$.

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B} = 3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple vector product $$\vec{A}\times\vec{B}\times\vec{C}$$.

Problem Statement

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B} = 3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple vector product $$\vec{A}\times\vec{B}\times\vec{C}$$.

$$\vec{A}\times\vec{B}\times\vec{C}=-7\hat{i}-\hat{j}+7\hat{k}$$

Problem Statement

Given $$\vec{A}=\hat{i}-2\hat{j}+2\hat{k}$$, $$\vec{B} = 3\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{C}=-\hat{i}-\hat{k}$$, calculate the triple vector product $$\vec{A}\times\vec{B}\times\vec{C}$$.

Solution

Let's work left to right.
$$\vec{A} \times \vec{B} = (\hat{i}-2\hat{j}+2\hat{k}) \times (3\hat{i}-\hat{j}-\hat{k}) =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -1 & -1 \end{vmatrix} =$$ $$(2-(-2))\hat{i} - (-1-6)\hat{j} + (-1-(-6))\hat{k} =$$ $$4\hat{i} + 7\hat{j} +5\hat{k} = \vec{v}$$
Now we need to calculate $$\vec{v} \times \vec{C}$$.
$$(4\hat{i} + 7\hat{j} +5\hat{k}) \times (-\hat{i}-\hat{k}) =$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 7 & 5 \\ -1 & 0 & -1 \end{vmatrix} =$$ $$(-7-0)\hat{i} - (-4-(-5))\hat{j} +(0-(-7))\hat{k} =$$ $$-7\hat{i} -\hat{j} +7\hat{k}$$

$$\vec{A}\times\vec{B}\times\vec{C}=-7\hat{i}-\hat{j}+7\hat{k}$$