# Deriving algebraic definition of cross product

• ChiralSuperfields
In summary: Homework Equation: ##\vec {u} \times \vec {v} = |u||v|sin(\theta)\hat n####\hat n = \frac{\vec u \times \vec v}{|\vec u \times \vec v|}##In summary, the equation for the vector cross product is defined as ##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j +
ChiralSuperfields
Homework Statement
I am trying to derive the formula ##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j + (u_1v_2 - u_2v_1)\hat k ## because it is extremely difficult to commit to memory for my upcoming exam
Relevant Equations
##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j + (u_1v_2 - u_2v_1)\hat k ##
So far, I have got the equations,
##u \cdot (\vec u \times \vec v) = 0##
##u_1a + u_2b + u_3c = 0##
##v_1a + v_2b + v_3c = 0##

Could some please give me some guidance?

Many thanks!

There used to be the indexing mnemonic xyzzy for x = yz - zy and the permutations of it yzxxz and zxyyx

But the easiest way is the determinant scheme

Code:
|    i    j    k  |
|   ax   ay   az  |
|   bx   by   bz  |

expanding the determinant over I j and k we get:

Code:
i | ay az |  - j | ax az | + k | ax ay |  = i (aybz-azby) - j (axbz-azbx) + k (axby-aybx)
| by bz |      | bx bz |     | bx by |

NOTE: Pay attention to the minus sign on the j term

Last edited:
WWGD, jtbell, topsquark and 3 others
Here's a video discussing it:

WWGD, topsquark and ChiralSuperfields
Alternatively, just remember $\mathbf{e}_z = \mathbf{e}_x \times \mathbf{e}_y$ together with the cyclic, anticommutative and distributive properties.

robphy, topsquark, ChiralSuperfields and 1 other person
ChiralSuperfields said:
I am trying to derive the formula ##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j + (u_1v_2 - u_2v_1)\hat k ## because it is extremely difficult to commit to memory for my upcoming exam
Why? There are some things that it's just better to memorize. Regarding the video that @jedishrfu posted, to evaluate a 3x3 determinant, you can break it down into three 2x2 determinants. It doesn't matter whether you go across a row or down a column. If you're expanding the "pseudo-determinant" of a cross product, it's probably simpler to expand across the top row.

##\begin{vmatrix} i & j & k\\a & b & c\\d & e & f \end{vmatrix}
=i\begin{vmatrix}b& c\\e & f\end{vmatrix} - j\begin{vmatrix}a& c\\d & f\end{vmatrix} + k \begin{vmatrix}a& b\\d & e\end{vmatrix}##
The important things to remember are:
1. Each 2x2 sub-determinant is made up of the terms not in the same row or column as the multiplier. For example, the first sub-determinant above consists of the terms of the 3x3 determinant that aren't in row 1 and column 1.
2. The signs of the multipliers of the sub-determinant alternate in sign. An easy way to remember is to look at the row and column numbers -- if they add up to an odd number, the sign should be -, otherwise it should be +. For example, the j entry is row 1, column 2. 1 + 2 = 3, which is odd, so the sign for the j multiplier should be -.
3. Evaluating a 2x2 determinant is straightforward. ##\begin{vmatrix}a & b\\c & d\end{vmatrix} = ad - bc##

topsquark and ChiralSuperfields
ChiralSuperfields said:
I am trying to derive the formula ##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j + (u_1v_2 - u_2v_1)\hat k ## because it is extremely difficult to commit to memory for my upcoming exam

It is a definition, not a derivation.

topsquark and ChiralSuperfields
I guess you could say it depends as the definition for vector cross product is usually defined as

##A \times B = |A||B|sin(\theta) \hat{n} ##

where ##\theta## is the angle between them and ##\hat{n}## is a unit vector that is perpendicular to both A and B vectors.

https://en.wikipedia.org/wiki/Cross_product

EDIT: As @Mark44 points out in a later post, I forgot to mention that the resultant vector is perpendicular to both A and B vectors.

Last edited:
topsquark and ChiralSuperfields
ChiralSuperfields said:
because it is extremely difficult to commit to memory for my upcoming exam
I learned the determinant method for memorizing the vector cross product, and it's always made it easy to remember how to calculate it. And if you're comfortable mentally wrapping around the diagonals, you don't need to write the extended determinant that is shown in the link below. Hope this helps.

https://efcms.engr.utk.edu/ef151-2019-01/pilot/classmgr.php?c=45&p=crossproduct

topsquark and ChiralSuperfields
Thank you all (@jedishrfu , @pasmith, @Mark44, @malawi_glenn , @berkeman ) for your help!

I understand now, we had not done the determinant in class yet (still a couple of weeks away after the test!), but now I can easily remember it :)

berkeman
Haha, we will test you in a few weeks!

ChiralSuperfields and berkeman
jedishrfu said:
I guess you could say it depends as the definition for vector cross product is usually defined as

##A \times B = |A||B|sin(\theta)## where ##\theta## is the angle between them

https://en.wikipedia.org/wiki/Cross_product
That isn't the formula for the cross product -- it's the magnitude of the cross product. IOW, for ##|A \times B|##. The wikipedia page gives this formula:

##a \times b = |a||b|sin(\theta) n##, where n is a unit vector perpendicular to both a and b.

malawi_glenn, topsquark, jedishrfu and 1 other person
ChiralSuperfields said:
Homework Statement: I am trying to derive the formula
##\vec {u} \times \vec {v} = (u_2v_3 - u_3v_2)\hat i + (u_3v_1 - u_1v_3)\hat j + (u_1v_2 - u_2v_1)\hat k ##

In a variant of your notation, I would write
##\vec {u} \times \vec {v} = \hat e_1(u_2v_3 - u_3v_2) + \hat e_2(u_3v_1 - u_1v_3) + \hat e_3(u_1v_2 - u_2v_1)##

Note the pattern...
123, 231, 312 (via cyclic permutations) are all even permutations of 123 ...and get the +
132, 321, 213 (via cyclic permutations) are all odd permutations of 123 ...and get the -

(Note this is seen in the post by @jedishrfu above,
as well as the posts by @Mark44 and
by @berkeman above. .. but keep the first factor in place in each term
(like what a simple computer would do to follow the pattern)
... don't rearrange (which would require more code for the computer to do).

The post by @pasmith has the most compact explanation.

I think the geometrical meaning should be kept in mind.
The cross-product is the oriented area of a parallelogram formed by the ordered pair of vetors ##(\vec u, \vec v)##.
In 3-dim, we can associate this area with a [pseudo]vector perpendicular to the plane, following the right-hand-rule. (In my opinion, the area feature is more important and more fundamental that the vector-via-the-right-hand-rule.)

Last edited:
ChiralSuperfields
robphy said:
In a variant of your notation, I would write
##\vec {u} \times \vec {v} = \hat e_1(u_2v_3 - u_3v_2) + \hat e_2(u_3v_1 - u_1v_3) + \hat e_3(u_1v_2 - u_2v_1)##

Note the pattern...
123, 231, 312 (via cyclic permutations) are all even permutations of 123 ...and get the +
132, 321, 213 (via cyclic permutations) are all odd permutations of 123 ...and get the -

(Note this is seen in the post by @jedishrfu above,
as well as the posts by @Mark44 and
by @berkeman above. .. but keep the first factor in place in each term
(like what a simple computer would do to follow the pattern)
... don't rearrange (which would require more code for the computer to do).

The post by @pasmith has the most compact explanation.

I think the geometrical meaning should be kept in mind.
The cross-product is the oriented area of a parallelogram formed by the ordered pair of vetors ##(\vec u, \vec v)##.
In 3-dim, we can associate this area with a [pseudo]vector perpendicular to the plane, following the right-hand-rule. (In my opinion, the area feature is more important and more fundamental that the vector-via-the-right-hand-rule.)
Thank you for sharing that @robphy!

• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Quantum Physics
Replies
12
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
839
• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Special and General Relativity
Replies
4
Views
827
• Linear and Abstract Algebra
Replies
22
Views
3K
• High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
• General Math
Replies
4
Views
4K
• Linear and Abstract Algebra
Replies
5
Views
2K