Cross Product
The cross product is a vector operation that acts on vectors in three dimensions and results in another vector in three dimensions. In contrast to dot product, which can be defined in both 2-d and 3-d space, the cross product is only defined in 3-d space. Another difference is that while the dot-product outputs a scalar quantity, the cross product outputs another vector.
Contents
Definition
The algebraic interpretation of the cross product is defined as
If \(\vec{a}=(a_{1},a_{2},a_{3})\) and \(\vec{b}=(b_{1},b_{2},b_{3})\) are 2 vectors in 3-d space, then the cross product \(\vec{a}\times \vec{b}\) can be defined in the following ways:
(A) as the determinant \[\begin{vmatrix} i & j & k \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 }\end{vmatrix} ,\] using the first row;
(B) as \(\left({ a }_{ 2 }{ b }_{ 3 }-{ a }_{ 3 }{ b }_{ 2 }, { a }_{ 3 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 3, }, { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 }\right)\) in vector notation. \(_\square\)
Find the cross product of \(\vec{v}=(3,6,8)\) and \(\vec{w}=(2,-4,7).\)
Method 1:
As the determinant \[\begin{align} \begin{vmatrix} i & j & k \\ 3 & 6 & 8 \\ 2 & -4 & 7 \end{vmatrix} =&(6(7)-8(-4))i-(3(7)-2(8))j+(3(-4)-(6)(2))k\\ =&74i-5j-24k\\ =&(74,-5,-24). \end{align}\]Method 2:
Simply using the formula from definition (B), we have: \[\begin{align} (v_2 w_3-v_{ 3 }{ w }_{ 2 }, { v }_{ 3 }{ w }_{ 1 }-{ v }_{ 1 }{ w }_{ 3, }, { v }_{ 1 }{ w }_{ 2 }-{ v }_{ 2 }{ w }_{ 1 }) &=(6(7)-8(-4), 2(8)-3(7), 3(-4)-(6)(2))\\ &=(74,-5,-24). \ _\square \end{align}\]
Find the cross product of the vectors \(\vec{v}=3i-2j-k\) and \(\vec{w}=4i-3j+2k.\)
We can get the cross product as the determinant of the following matrix: \[\begin{align} \begin{vmatrix} i & j & k \\ 3 & -2 & -1 \\ 4 & 3 & 2 \end{vmatrix} &=(-2(2)-3(-1))i-(3(2)-4(-1))j+(3(3)-4(-2))k\\ &=-i-10j+17k.\ _\square \end{align}\]
Properties
Some of the main properties of the cross product are as follows:
a) The cross product is not commutative. That is, \(\vec{a}\times \vec{b} \neq \vec{b}\times \vec{a}\) . By the right-hand screw rule, \(\vec{a}\times \vec{b}\) and \(\vec{b}\times \vec{a}\) have opposite directions, so \(\vec{a}\times \vec{b} = -\vec{b}\times \vec{a}\).
b) The cross products are distributive with respect to vector addition: \(\vec{a}\times (\vec{b}+ \vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c} \).
c) \(\ 0\times \vec{a} = \vec{a}\times 0=0\).
d) \(\ c(\vec{a}\times \vec{b})=(c\vec{a})\times \vec{b}=\vec{a}\times (c\vec{b})\).
e) \(\ \vec{a}\times \vec{a}=0\).
f) Under reflection (that is, taking mirror image), all components of a vector change signs. Thus we have \(\vec{a}\times \vec{b} \rightarrow (-\vec{a})\times (-\vec{b})=\vec{a}\times \vec{b}\). Thus, \(\vec{a}\times \vec{b}\) remains the same under reflection. \(_\square\)
All these properties can are derived from the definition of the cross product and are left as exercises for the readers to prove.
Given two vectors \(\vec{v}=3i-2j-k\) and \(\vec{w}=4i-3j+2k\), find
A) \(\ \vec{v}\times \vec{w}\)
B) \(\ \vec{w}\times \vec{v}\).
A) \(\ \vec{v}\times \vec{w}:\)
We can get the cross product by calculating the determinant: \[\begin{align} \begin{vmatrix} i & j & k \\ 3 & -2 & -1 \\ 4 & 3 & 2 \end{vmatrix} &=(-2(2)-3(-1))i-(3(2)-4(-1))j+(3(3)-4(-2))k\\ &=-i-10j+17k. \end{align}\]B) \(\ \vec{w}\times \vec{v}:\)
We could just follow the same steps, but a shortcut would be to use theorem a) above: \[\begin{align} \vec{w}\times \vec{v}&=-(\vec{v}\times \vec{w})\\ &=-(-i-10j+17k)\\ &=i+10j-17k.\ _\square \end{align}\]
There are also some properties that relate the cross product and the dot product.
A) \(\ \vec{a}\cdot (\vec{a}\times \vec{b})=0\)
B) \(\ \vec{b}\cdot (\vec{a}\times \vec{b})=0\)
C) \(\ \vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}\)
D) \(\ (\vec{a}\times \vec{b})\times \vec{c}=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}\)
E) \(\ { \left\| \vec { a } \times \vec { b } \right\| }^{ 2 }=\left\| \vec { a } \right\|^2 \left\| \vec { b } \right\|^2 -{ (\vec { a } \cdot \vec { b } ) }^{ 2 } \qquad\) (known as Lagrange's identity) \(_\square\)
The first two properties are easy to understand if we realize that the cross product outputs a vector perpendicular to both vectors, and that the dot product of perpendicular vectors is zero. The others are like the first two as a result of the properties of the dot and cross products, and are left for the reader to prove.
Given the vectors \(\vec{a}=(-1,2,2)\), \(\vec{b}=(0,3,4)\) and \(\vec{c}=(1,-2,0)\), show that they do indeed satisfy the above four properties A) through D).
A) \(\ \vec{a}\cdot (\vec{a}\times \vec{b}):\)
\[\begin{align} \vec { a } \times \vec { b } =\begin{bmatrix} i & j & k \\ -1\quad & 2\quad & 2\quad \\ 0 & 3 & 4 \end{bmatrix} =(2,4,-3) \implies \vec { a } \cdot (\vec { a } \times \vec { b } ) =(-1)(2)+2(4)+2(-3) =0. \end{align}\]B) \(\ \vec{b}\cdot (\vec{a}\times \vec{b}):\)
\[\begin{align} \vec { a } \times \vec { b } =\begin{bmatrix} i & j & k \\ -1\quad & 2\quad & 2\quad \\ 0 & 3 & 4 \end{bmatrix} =(2,4,-3) \implies \vec { b } \cdot (\vec { a } \times \vec { b } ) =0(2)+3(4)+4(-3) =0. \end{align}\]C) \(\ \vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}:\)
We have \(\vec{a}\times (\vec{b}\times \vec{c})=(-14,13,-20)\), which will be left for the readers to show how. Then \[\begin{align} \vec { a } \cdot \vec { c } &=-1(1)+2(-2)+2(0)\\ &=-5\\
\vec { a } \cdot \vec { b } &=-1(0)+2(3)+2(4)\\ &=14\\ \vec { a } \times (\vec { b } \times \vec { c } ) &=(\vec { a } \cdot \vec { c } )\vec { b } -(\vec { a } \cdot \vec { b } )\vec { c } \\ \Rightarrow (-14,13,-20) &=-5(0,3,4)-14(1,-2,0)\\ &=(0,15,-20)-(14,-28,0)\\ &=(-14,13,-20). \end{align}\]D) \(\ (\vec{a}\times \vec{b})\times \vec{c}=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}:\) \[\begin{align} \vec { a } \cdot \vec { c } &=-1(1)+2(-2)+2(0)\\ &=-5\\ \vec { b } \cdot \vec { c } &=0(1)+3(-2)+4(0)\\ &=-6\\ (\vec { a } \times \vec { b } )\times \vec { c } &=(\vec { a } \cdot \vec { c } )\vec { b } -(\vec { b } \cdot \vec { c } )\vec { a } \\ \Rightarrow (-6,-3,-8) &=-5(0,3,4)-(-6)(-1,2,2)\\ &=(0,-15,20)-(6,-12,-12)\\ &=(-6,-3,-8).\ _\square \end{align}\]
Geometric Definition
Consider three-dimensional vectors \(\vec{a}\) and \(\vec{b}\), and let \(\theta\) be the angle between them . The geometric interpretation of the cross product is a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) (using the right-hand rule) and has norm defined as
\[\left\| a\times b \right\| =\left\| \vec { a } \right\| \left\| \vec { b } \right\| \sin\theta.\ _\square\]
Given two vectors \(\vec{v}=3i+2j-0k\) and \(\vec{w}=5i-j+0k,\) what is the magnitude of the cross product if the angle between the two vectors is \( { 45 }^{ \circ }\)?
We have: \[\begin{align} \left\|\vec{ v} \right\| &=\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }-{ 0 }^{ 2 } } =\sqrt { 13 } \\ \left\|\vec{ w }\right\| &=\sqrt { { 5 }^{ 2 }+{ (-1 })^{ 2 }+{ 0 }^{ 2 } } =\sqrt { 26 } . \end{align}\] Then using the definition for magnitude of a cross product, we have \[\left\| \vec { v } \times \vec { w } \right\| =\sqrt { 26 } \cdot \sqrt { 13 }\cdot \frac { \sqrt { 2 } }{ 2 } =\frac { 26 }{ 2 } =13.\ _\square\]
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 \(\vec{a}\) and \(\vec{b}\) . This is given in the definition of the geometric interpretation. For the algebraic interpretation, taking the dot product gives:
\[\begin{align} (a_1, a_2, a_3)\cdot (a_2b_3-a_3b_3, { a }_{ 3 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 3, }, { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 })&=0\\ ({ b }_{ 1 }{ b }_{ 2 }{ ,b }_{ 3 })\cdot ({ a }_{ 2 }{ b }_{ 3 }-{ a }_{ 3 }{ b }_{ 3 }, { a }_{ 3 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 3, }, { a }_{ 1 }{ b }_{ 2 }-{ a }_{ 2 }{ b }_{ 1 })&=0. \end{align}\]
Furthermore, the vectors point in the same direction since the determinant
\[\begin{vmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { a }_{ 2 }{ b }_{ 3 }{ -a }_{ 3 }{ b }_{ 3 } & { a }_{ 3 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 3 } & { a }_{ 1 }{ b }_{ 2 }{ -a }_{ 2 }{ b }_{ 1 } \end{vmatrix}\]
is positive by expanding along the third row; and 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
\[\begin{align} \Vert \vec{a} \Vert^2 \Vert \vec{b} \Vert^2 \sin^2 \theta =& (a_1 ^2 + a_2 ^2 + a_3 ^2 )(b_1 ^2 + b_2 ^2 + b_3 ^2 )\left(1 - \left( \frac { (a_1b_1 + a_2b_2 + a_3 b_3) }{ \sqrt{ (a_1 ^2 + a_2 ^2 + a_3 ^2 )(b_1 ^2 + b_2 ^2 + b_3 ^2 ) }} \right) ^2\right) \\ =& a_1^2 b_2 ^2 + a_1 ^2 b_3 ^2 + a_2 ^2 b_3 ^2 + a_2 ^2 b_1 ^2 + a_3 ^2 b_1 ^2 + a_3 ^2 b_2 ^2 -2a_1a_2b_1b_2 - 2 a_2a_3 b_2b_3 - 2a_3a_1b_3b_1. \end{align}\]
For the algebraic interpretation, we have
\[\begin{align} & \left(a_2b_3-b_2a_3\right)^2 + \left(-a_1b_3 + b_1a_3\right)^2 + \left(a_1b_2-b_1a_2 \right)^2 \\ =& \left(a_2^2 b_3^2 + b_2 ^2 a_3 ^2 - 2 a_2b_2 a_3 b_3\right) + \left(a_1^2 b_3^2 + a_3^2b_1^2 - 2a_3a_1b_3 b_1\right) + \left(a_1^2 b_2^2 + b_1^2 a_2^2 - 2a_1a_2b_1b_2\right). \end{align} \]
By comparing terms, we see that the lengths of the vectors are equal. \(_\square\)
We know that the area of a parallelogram is given by \((\text{base}) \times (\text{height}).\) In the figure above, the area is the product of the magnitudes of \(\vec{a}\) and \(\vec{b}:\)
\[\text{Area}=\left\| \vec { a } \right\| \left\| \vec { b } \right\| \sin\theta .\]
We can see that the area of the parallelogram is equal to the magnitude of the cross product \(\left(\left\| \vec { a } \times \vec { b } \right\| =\left\| \vec { a } \right\| \left\| \vec { b } \right\| \sin\theta \right).\) That is, the magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors.
Find the area of the triangle formed by the vectors \(A=(1,0,-1)\), \(B=(2,1,1)\), and \(C=(-1,2,1).\)
We have
\[\begin{align} \vec{AB}&=(1,1,2)\\ \vec{AC}&=(-2,2,2). \end{align}\]
Since the cross product of \(AB\) and \(AC\) is the area of the parallelogram, the area of the triangle is equal to half the area of the parallelogram:
\[\begin{align} A_{T}&=\frac{1}{2}\left\| \vec { AB } \times \vec { AC } \right\|\\ \\ \vec { AB } \times \vec { AC } &=\begin{bmatrix} i & j & k \\ 1 & 1 & 2 \\ -2\quad & 2\quad & 2\quad \end{bmatrix}\\ &=(2-4)i-(2-(-4))j+(2-(-2))k\\ &=-2i-6j+4k\\ \\ \left\| \vec { AB } \times \vec { AC } \right\| &=\sqrt { { (-2 })^{ 2 }+{ (-6) }^{ 2 }+{ (4) }^{ 2 } }\\ &=\sqrt { 56 }\\ \\ \Rightarrow A_{ T }&=\frac { \sqrt { 56 } }{ 2 }=\sqrt{14}. \ _\square \end{align} \]
Show that the cross product \(\vec{a}\cdot ( \vec{b}\times \vec{c})=\vec{b}\cdot ( \vec{c}\times \vec{a})=\vec{c}\cdot ( \vec{a}\times \vec{b})\) is equal to the volume of of the parallelepiped.
From the properties of the cross product, the area of the base of the parallelepiped is
\[\text{A}_{\text{base}}=\left\| \vec { a } \times \vec { b } \right\| .\]
Let \(\theta\) be the angle formed by the vector \(\vec{c}\) with the vertical, then the height of the parallelpiped is
\[\text{Height}=\left\| \vec { c } \right\|cos\theta. \]
Now, since \( \vec { a } \times \vec { b }\) is a vector normal to the plane formed by vectors \(\vec{a}\) and \(\vec{b}\), \(\theta\) is also the angle between vectors \(\vec{a}\times \vec{b}\) and \(\vec{c}\). So the area of the parallelpiped is
\[\begin{align} {\text{A}} &=\text{A}_{\text{base}} \times \text{Height}\\ &=\left\| \vec { a } \times \vec { b } \right\| \left\| \vec { c } \right\| cos\theta \\ &=(\vec { a } \times \vec { b } )\cdot \vec { c }. \ _\square \end{align} \]