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, 4 months ago

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