## 17Calculus - Vector Cross Product

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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

Practice

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

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

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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

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

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

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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?

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

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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 video

video by Krista King Math

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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 video

video by Krista King Math

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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 video

video by Krista King Math

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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

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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.

### Animations for Physics and Astronomy - 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.

Practice

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.

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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.

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

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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 an 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.

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

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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.

 Triple Scalar Product Triple Vector Product

### dot and cross products 17calculus youtube playlist

You CAN Ace Calculus

 basics of vectors dot product linear algebra

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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