Cross Product of Vectors

The cross product is an operation that acts on vectors in three dimensions and results in another vector in three dimensions. In contrast to the dot product, the cross product is restricted to vectors in three dimensions. Consider three-dimensional vectors a=[a1,a2,a3]\vec{a} = [a_1, a_2, a_3] and b=[b1,b2,b3]\vec{b} = [b_1, b_2, b_3 ] and let θ \theta be the angle between a \vec{a} and b \vec{b} . The geometric interpretation of the cross product is a vector a×b \vec{a} \times \vec{b} that is perpendicular to both a \vec{a} and b \vec{b} (using the right hand rule), and has norm absinθ \lVert \vec{a} \rVert \lVert \vec{b} \rVert \sin \theta .

The algebraic interpretation of the cross product is a vector obtained by the formula:

a×b=[a1,a2,a3]×[b1,b2,b3]=[a2b3b2a3,a1b3+b1a3,a1b2b1a2]. \vec{a} \times \vec{b} = [a_1 , a_2 , a_3 ] \times [b_1, b_2, b_3 ] = [a_2b_3-b_2a_3, -a_1b_3+b_1a_3, a_1b_2-b_1a_2].

This formula can be expressed as the determinant of a matrix:

[a1,a2,a3]×[b1,b2,b3]=det(a1a2a3b1b2b3ijk). [a_1 , a_2 , a_3 ] \times [b_1, b_2, b_3 ] = \det \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \vec{i} & \vec{j} & \vec{k} \\ \end{pmatrix}.

Now, we can use various properties about determinants to prove properties of the cross product. For example, we can verify that the cross product is distributive over addition, i.e.,

a×(b+c)=a×b+a×c. \vec{a}\times (\vec{b}+\vec{c}) =\vec{a}\times\vec{b} + \vec{a}\times \vec{c} .

The cross product satisfies the following properties:

  1. Since sinθ \sin \theta is an odd function, we have a×b=b×a \vec{a} \times\vec{b} = -\vec{b} \times \vec{a} .

  2. What happens if b=a \vec{b} = \vec{a} ? We can no longer determine the unique direction of a vector that is perpendicular to a \vec{a} (remember that we are in three dimensions), and the geometric interpretation seems unclear. However, a×a \vec{a} \times \vec{a} has norm aasin(0)=0 \lVert \vec{a} \rVert \lVert \vec{a} \rVert \sin (0) = 0 and thus a×a=0 \vec{a} \times \vec{a}= \vec{0} .

  3. If a×b=0 \vec{a}\times\vec{b} = \vec{0} , what can we say about a \vec{a} and b \vec{b} ? By definition, a×b=0 \vec{a}\times\vec{b} = \vec{0} implies absinθ=0 \Vert \vec{a} \Vert \Vert \vec{b} \Vert \sin \theta = 0 . Therefore,

a×b=0 if and only if a=0 or b=0 or a is parallel to b. \vec{a}\times\vec{b} = \vec{0} \mbox{ if and only if } \vec{a}=\vec{0} \mbox{ or } \vec{b}=\vec{0} \mbox{ or } \vec{a} \mbox{ is parallel to } \vec{b} .

Worked Examples

1. Show that a×b \Vert \vec{a}\times \vec{b} \Vert is the positive area of the parallelogram having sides a \vec{a} and b \vec{b} .

Solution: We know the area of the triangle bounded by vectors a \vec{a} and b \vec{b} is given by 12absinθ \frac {1}{2} \Vert \vec{a} \Vert \Vert \vec{b} \Vert \sin \theta . Hence, the area of the parallelogram is absinθ \Vert \vec{a} \Vert \Vert \vec{b} \Vert \sin \theta , which is the norm of a×b \vec{a}\times \vec{b} , i.e. a×b \Vert \vec{a} \times\vec{b} \Vert .

It is interesting that while area is a two-dimensional property, this uses a three-dimensional calculation.

 

2. Show that the cancellation law doesn't hold. If a×b=a×c \vec{a}\times\vec{b} =\vec{a}\times \vec{c} and a0 \vec{a}\neq 0 , what can we say about the vectors a,b,c \vec{a}, \vec{b}, \vec{c} ?

We have that a×(bc)=0 \vec{a}\times (\vec{b}-\vec{c}) = 0 . From the remark, since a0 \vec{a}\neq 0 , we either have bc=0 \vec{b}-\vec{c} = 0 or a \vec{a} is parallel to bc \vec{b}-\vec{c} . In the first case, we get b=c \vec{b}=\vec{c}, and in the second case, we get c=b+ka \vec{c} =\vec{b} + k\vec{a} for some value kR k \in \mathbb{R} .

 

3. Prove Lagrange's Identity. a×b2+(ab)2=a2b2 \Vert \vec{a}\times \vec{b} \Vert ^2 + (\vec{a} \cdot \vec{b})^2 = \Vert \vec{a} \Vert^2 \Vert \vec{b} \Vert^2

Solution. This follows immediately, since a×b2=(absinθ)2 \Vert \vec{a} \times \vec{b} \Vert ^2 = (\Vert \vec{a} \Vert\Vert \vec{b} \Vert\sin \theta)^2 , (ab)2=(abcosθ)2 (\vec{a} \cdot \vec{b} )^2 = (\Vert \vec{a} \Vert\Vert \vec{b} \Vert\cos \theta)^2 , and cos2θ+sin2θ=1 \cos^2 \theta + \sin ^2 \theta = 1 .

 

4. Show the equivalence of the geometric interpretation and the algebraic interpretation of the cross product.

We need to show that the geometric and algebraic definitions give vectors with the same magnitude and direction. To check direction, we will show that both vectors are perpendicular to a \vec{a} and b\vec{b} . This is given in the definition of the geometric interpretation. For the algebraic interpretation, taking the dot product gives

[a1,a2,a3][a2b3b2a3,a1b3+b1a3,a1b2b1a2]=0, [a_1, a_2, a_3] \cdot [a_2b_3-b_2a_3, -a_1b_3+b_1a_3, a_1b_2-b_1a_2] = 0, [b1,b2,b3][a2b3b2a3,a1b3+b1a3,a1b2b1a2]=0. [b_1, b_2, b_3] \cdot [a_2b_3-b_2a_3, -a_1b_3+b_1a_3, a_1b_2-b_1a_2] = 0.

Furthermore, the vectors point in the same direction since the determinant of

(a1a2a3b1b2b3a2b3b2a3a1b3+b1a3a1b2b1a2) \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ a_2b_3-b_2a_3 & -a_1b_3+b_1a_3 & a_1b_2-b_1a_2 & \\ \end{pmatrix}

is positive by expanding along the third row; hence the vectors are determined by the right hand rule.

Next, we check that the vectors have the same length, by calculating the square of the norm. We have

a2b2sin2θ=(a12+a22+a32)(b12+b22+b32)(1((a1b1+a2b2+a3b3)(a12+a22+a32)(b12+b22+b32))2)=a12b22+a12b32+a22b32+a22b12+a32b12+a32b222a1a2b1b22a2a3b2b32a3a1b3b1\begin{aligned}{ } & \Vert \vec{a} \Vert^2 \Vert \vec{b} \Vert^2 \sin^2 \theta \\ & = (a_1 ^2 + a_2 ^2 + a_3 ^2 )(b_1 ^2 + b_2 ^2 + b_3 ^2 )\left(1 - \left( \frac { (a_1b_1 + a_2b_2 + a_3 b_3) }{ \sqrt{ (a_1 ^2 + a_2 ^2 + a_3 ^2 )(b_1 ^2 + b_2 ^2 + b_3 ^2 ) }} \right) ^2\right) \\ & = a_1^2 b_2 ^2 + a_1 ^2 b_3 ^2 + a_2 ^2 b_3 ^2 + a_2 ^2 b_1 ^2 + a_3 ^2 b_1 ^2 + a_3 ^2 b_2 ^2 -2a_1a_2b_1b_2 - 2 a_2a_3 b_2b_3 - 2a_3a_1b_3b_1 \end{aligned}

For the algebraic interpretation, we have

(a2b3b2a3)2+(a1b3+b1a3)2+(a1b2b1a2)2=(a22b32+b22a322a2b2a3b3)+(a12b32+a32b122a3a1b3b1)+(a12b22+b12a222a1a2b1b2).\begin{aligned} (a_2b_3-b_2a_3)^2 & + (-a_1b_3 + b_1a_3)^2 + (a_1b_2-b_1a_2 )^2 \\ &= (a_2^2 b_3^2 + b_2 ^2 a_3 ^2 - 2 a_2b_2 a_3 b_3) + (a_1^2 b_3^2 + a_3^2b_1^2 - 2a_3a_1b_3 b_1) \\ & + (a_1^2 b_2^2 + b_1^2 a_2^2 - 2a_1a_2b_1b_2). \end{aligned}

By comparing terms, we see that the lengths of the vectors are equal.

The cross product has numerous applications in physics, such as describing moments, angular momentum, torque, etc.

Note by Arron Kau
5 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...