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You CAN Ace Calculus

17calculus > vectors > cross product

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

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Vector 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
 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

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

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

 Larson Calculus - proofs

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

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
 Cross Product Applications

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

The triple scalar product is so named because the result is a scalar. [For comparison, see the triple vector product panel 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
 Dr Chris Tisdell - Scalar triple product and volume
 Larson Calculus -Proof - Geometric Property of the Triple Scalar Product

### 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 $$\vec{u} \times \vec{v} \times \vec{w}$$.

### 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]

### Search 17Calculus

Practice Problems

 Level A - Basic

Practice A01

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

Practice A02

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

Practice A03

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

Practice A04

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

Practice A05

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

Practice A06

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

Practice A07

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

Practice A08

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

solution

Practice A09

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

Practice A10

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

Practice A11

Calculate the cross product of the vectors $$\vec{a}=\langle5,-1,-2\rangle$$ and $$\vec{b}=\langle-3,2,4\rangle$$.

solution

Practice A12

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

Practice A13

Calculate the cross product of the vectors $$\vec{a}=\langle2,-3\rangle$$ and $$\vec{b}=\langle4,5\rangle$$.

solution

Practice A14

Calculate the cross product of the vectors $$\vec{a}=\langle5,1,4\rangle$$ and $$\vec{b}=\langle-1,0,2\rangle$$.

solution

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