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.
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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.
\(\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = \) \( (u_2 v_3 - u_3 v_2)\hat{i} - (u_1 v_3 - u_3 v_1)\hat{j} + (u_1 v_2 - u_2 v_1)\hat{k} \)
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 |
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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 |
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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 |
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\( (2\hat{i} + 3\hat{j} + 4\hat{k} ) \times ( \hat{i} + 3\hat{k} ) = 9\hat{i} - 2\hat{j} - 3\hat{k} \)
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
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 0 & 3 \end{vmatrix} = \) \((9-0)\hat{i} - (6-4)\hat{j} + (0-3)\hat{k} = \) \( 9\hat{i} - 2\hat{j} - 3\hat{k} \)
Final Answer
\( (2\hat{i} + 3\hat{j} + 4\hat{k} ) \times ( \hat{i} + 3\hat{k} ) = 9\hat{i} - 2\hat{j} - 3\hat{k} \)
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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 |
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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 \(\vec{A}\times\vec{B}\).
Final Answer |
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\( \vec{A} \times \vec{B} = 4\hat{i} + 7\hat{j} + 5\hat{k} \)
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
\(\vec{A} \times \vec{B} = (\hat{i}-2\hat{j}+2\hat{k}) \times (3\hat{i}-\hat{j}-\hat{k}) = \) \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -1 & -1 \end{vmatrix} = \) \((2 -(-2))\hat{i} - (-1-6)\hat{j} + (-1-(-6))\hat{k} = \) \( 4\hat{i} + 7\hat{j} +5\hat{k} \)
Final Answer
\( \vec{A} \times \vec{B} = 4\hat{i} + 7\hat{j} + 5\hat{k} \)
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Calculate the cross product of \(\hat{i}\) and \(\hat{j}\).
Problem Statement |
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Calculate the cross product of \(\hat{i}\) and \(\hat{j}\).
Final Answer |
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\(\hat{i}\times\hat{j}=\hat{k}\)
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?
Final Answer
\(\hat{i}\times\hat{j}=\hat{k}\)
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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
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
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
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.
video by Larson Calculus |
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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 |
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Here is a video with proofs of some of these geometric properties.
video by Larson Calculus |
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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 |
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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 |
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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 |
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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 |
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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.
\( \| \vec{A} \times \vec{B} \| \) |
\( \| 9\hat{i} - 2\hat{j} - 3\hat{k} \| \) |
\( \sqrt{9^2 + (-2)^2 + (-3)^2} \) |
\( \sqrt{81+4+9} = \sqrt{94} \) |
\( \|\vec{A} \| = \sqrt{2^2+3^2+4^2} = \sqrt{4+9+16} = \sqrt{29}\)
\( \|\vec{B}\| = \sqrt{1^2+0^2+3^2} = \sqrt{10}\)
\(\displaystyle{ \sin(\theta) = \frac{\| \vec{A} \times \vec{B} \|}{\|\vec{A}\| \|\vec{B}\|} }\) |
\(\displaystyle{ \sin(\theta) = \frac{\sqrt{94}}{\sqrt{29}\sqrt{10}} }\) |
\(\displaystyle{ \theta = \arcsin\left( \frac{\sqrt{94}}{\sqrt{290}} \right) \approx 0.606 }\) |
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.
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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 |
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Find a unit vector that is perpendicular to \( \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{B} = 3\hat{i} + 2\hat{j} + \hat{k} \).
Final Answer |
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\(\displaystyle{ \vec{u} = \frac{\pm 1}{\sqrt{6}} (-\hat{i}+2\hat{j}-\hat{k}) }\)
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.
Final 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 |
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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.
Final Answer |
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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 \( \|\vec{A}\| \|\vec{B}\| \).
\( \|\vec{A}\| \|\vec{B}\| \) |
\( \| 2\hat{i}+3\hat{j}+6\hat{k} \| \| 6\hat{i} + 2 \hat{j}-3\hat{k} \| \) |
\( \sqrt{2^2 + 3^2 + 6^2} \sqrt{6^2 + 2^2 + (-3)^2} \) |
\( \sqrt{4 + 9 + 36} \sqrt{36 +4 + 9} \) |
\( \sqrt{49}\sqrt{49} = 7(7) = 49 \) |
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.
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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 |
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Find a unit vector that is perpendicular to \( \vec{A} = 2\hat{i} + 3\hat{j} + 6\hat{k} \) and \( \vec{B} = 6\hat{i} + 2\hat{j} - 3\hat{k} \).
Final Answer |
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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.
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 \(\displaystyle{\frac{\pm 1}{7}[3\hat{i}-6\hat{j}+2\hat{k}]}\).
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Let A, B, C and D be the four corners of a parallelogram with point C 'opposite' from A. Given that \(A=(3,1,-2)\), \(B=(4,1,5)\) and \(D=(1,4,3)\), determine the location of C. Also, find the area of the parallelogram ABCD.
Problem Statement
Let A, B, C and D be the four corners of a parallelogram with point C 'opposite' from A. Given that \(A=(3,1,-2)\), \(B=(4,1,5)\) and \(D=(1,4,3)\), determine the location of C. Also, find the area of the parallelogram ABCD.
Solution
video by Steve Butler |
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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.
Really UNDERSTAND Calculus
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