Taking two vectors, we can write every combination of components in a grid:

This completed grid is the *outer product*, which can be separated into the:

**Dot product**, the interactions between similar dimensions (`x*x`

,`y*y`

,`z*z`

)**Cross product**, the interactions between different dimensions (`x*y`

,`y*z`

,`z*x`

, etc.)

The dot product ($\vec{a} \cdot \vec{b}$) measures similarity because it only accumulates interactions in matching dimensions. It’s a simple calculation with 3 components.

The cross product (written $\vec{a} \times \vec{b}$) has to measure a half-dozen “cross interactions”. The calculation looks complex but the concept is simple: accumulate 6 individual differences for the total difference.

Instead of thinking “When do I need the cross product?” think “When do I need interactions between different dimensions?”.

Area, for example, is formed by vectors pointing in different directions (the more orthogonal, the better). Indeed, the cross product measures the area spanned by two 3d vectors (source):

(The “cross product” assumes 3d vectors, but the concept extends to higher dimensions.)

Did the key intuition click? Let’s hop into the details.

## Defining the Cross Product

The dot product represents the similarity between vectors as a single number:

For example, we can say that North and East are 0% similar since $(0, 1) \cdot (1, 0) = 0$. Or that North and Northeast are 70% similar ($\cos(45) = .707$, remember that trig functions are percentages.) The similarity shows the amount of one vector that “shows up” in the other.

Should the cross product, the difference between vectors, be a single number too?

Let’s try. Sine is the percentage difference, so we could write:

Unfortunately, we’re missing some details. Let’s say we’re looking down the x-axis: both y and z point 100% away from us. A number like “100%” tells us there’s a big difference, but we don’t know what it is! We need extra information to tell us “the difference between $\vec{x}$ and $\vec{y}$ is *this*” and “the difference between $\vec{x}$ and $\vec{z}$ is *that*“.

So, let’s express the cross product as a vector:

The

*size*of the cross product is the numeric “amount of difference” (with $\sin(\theta)$ as the percentage). By itself, this doesn’t distinguish $\vec{x} \times \vec{y}$ from $\vec{x} \times \vec{z}$.The

*direction*of the cross product is based on both inputs: it’s the direction orthogonal to both (i.e., favoring neither).

Now $\vec{x} \times \vec{y}$ and $\vec{x} \times \vec{z}$ have different results, each with a magnitude indicating they are “100%” different from $\vec{x}$.

(Should the dot product be a vector result too? Well, we’re tracking the similarity between $\vec{a}$ and $\vec{b}$. The similarity measures the overlap between the original vector directions, which we already have.)

## Geometric Interpretation

Two vectors determine a plane, and the cross product points in a direction different from both (source):

Here’s the problem: there’s two perpendicular directions. By convention, we assume a “right-handed system” (source):

If you hold your first two fingers like the diagram shows, your thumb will point in the direction of the cross product. I make sure the orientation is correct by sweeping my first finger from $\vec{a}$ to $\vec{b}$. With the direction figured out, the magnitude of the cross product is $|a| |b| \sin(\theta)$, which is proportional to the magnitude of each vector and the “difference percentage” (sine).

## The Cross Product For Orthogonal Vectors

To remember the right hand rule, write the `xyz`

order twice: `xyzxyz`

. Next, find the pattern you’re looking for:

`xy => z`

(`x`

cross`y`

is`z`

)`yz => x`

(`y`

cross`z`

is`x`

; we looped around:`y`

to`z`

to`x`

)`zx => y`

Now, `xy`

and `yx`

have opposite signs because they are forward and backward in our `xyzxyz`

setup.

So, without a formula, you should be able to calculate:

Again, this is because `x`

cross `y`

is positive `z`

in a right-handed coordinate system. I used unit vectors, but we could scale the terms:

## Calculating The Cross Product

A single vector can be decomposed into its 3 orthogonal parts:

When the vectors are crossed, each pair of orthogonal components (like $a_x \times b_y$) casts a vote for where the orthogonal vector should point. 6 components, 6 votes, and their total is the cross product. (Similar to the gradient, where each axis casts a vote for the direction of greatest increase.)

`xy => z`

and`yx => -z`

(assume $\vec{a}$ is first, so`xy`

means $a_x b_y$)`yz => x`

and`zy => -x`

`zx => y`

and`xz => -y`

`xy`

and `yx`

fight it out in the `z`

direction. If those terms are equal, such as in $(2, 1, 0) \times (2, 1, 1)$, there is no cross product component in the `z`

direction (2 – 2 = 0).

The final combination is:

where $\vec{n}$ is the unit vector normal to $\vec{a}$ and $\vec{b}$.

Don’t let this scare you:

- There’s 6 terms, 3 positive and 3 negative
- Two dimensions vote on the third (so the
`z`

term must only have`y`

and`x`

components) - The positive/negative order is based on the
`xyzxyz`

pattern

If you like, there is an algebraic proof, that the formula is both orthogonal and of size $|a| |b| \sin(\theta)$, but I like the “proportional voting” intuition.

## Example Time

Again, we should do simple cross products in our head:

Why? We crossed the `x`

and `y`

axes, giving us `z`

(or $\vec{i} \times \vec{j} = \vec{k}$, using those unit vectors). Crossing the other way gives $-\vec{k}$.

Here’s how I walk through more complex examples:

- Let’s do the last term, the z-component. That’s (1)(5) minus (4)(2), or 5 – 8 = -3. I did
`z`

first because it uses`x`

and`y`

, the first two terms. Try seeing (1)(5) as “forward” as you scan from the first vector to the second, and (4)(2) as backwards as you move from the second vector to the first. - Now the
`y`

component: (3)(4) – (6)(1) = 12 – 6 = 6 - Now the
`x`

component: (2)(6) – (5)(3) = 12 – 15 = -3

So, the total is $(-3, 6, -3)$ which we can verify with Wolfram Alpha.

In short:

- The cross product tracks all the “cross interactions” between dimensions
- There are 6 interactions (2 in each dimension), with signs based on the
`xyzxyz`

order

## Appendix

**Connection with the Determinant**

You can calculate the cross product using the determinant of this matrix:

There’s a neat connection here, as the determinant (“signed area/volume”) tracks the contributions from orthogonal components.

There are theoretical reasons why the cross product (as an orthogonal vector) is only available in 0, 1, 3 or 7 dimensions. However, the cross product as a single number is essentially the determinant (a signed area, volume, or hypervolume as a scalar).

**Connection with Curl**

Curl measures the twisting force a vector field applies to a point, and is measured with a vector perpendicular to the surface. Whenever you hear “perpendicular vector” start thinking “cross product”.

We take the “determinant” of this matrix:

Instead of multiplication, the interaction is taking a partial derivative. As before, the $\vec{i}$ component of curl is based on the vectors and derivatives in the $\vec{j}$ and $\vec{k}$ directions.

**Relation to the Pythagorean Theorem**

The cross and dot product are like the orthogonal sides of a triangle:

For unit vectors, where $|a| = |b| = 1 $, we have:

I cheated a bit in the grid diagram, as we have to track the squared magnitudes (as done in the Pythagorean Theorem).

**Advanced Math**

The cross product & friends get extended in Clifford Algebra and Geometric Algebra. I’m still learning these.

**Cross Products of Cross Products**

Sometimes you’ll have a scenario like:

First, the cross product isn’t associative: order matters.

Next, remember what the cross product is doing: finding orthogonal vectors. If any two components are parallel ($\vec{a}$ parallel to $\vec{b}$) then there are no dimensions pushing on each other, and the cross product is zero (which carries through to $0 \times \vec{c}$).

But it’s ok for $\vec{a}$ and $\vec{c}$ to be parallel, since they are never directly involved in a cross product, for example:

Whoa! How’d we get back to $\vec{j}$? We asked for a direction perpendicular to both $\vec{i}$ and $\vec{j}$, and made that direction perpendicular to $\vec{i}$ again. Being “doubly perpendicular” means you’re back on the original axis.

**Dot Product of Cross Products**

Now if we take

what happens? We’re forced to do $\vec{a} \times \vec{b}$ first, because $\vec{b} \cdot \vec{c}$ returns a scalar (single number) which can’t be used in a cross product.

If $\vec{a}$ and $\vec{c}$ are parallel, what happens? Well, $\vec{a} \times \vec{b}$ is perpendicular to $\vec{a}$, which means it’s perpendicular to $\vec{c}$, so the dot product with $\vec{c}$ will be zero.

I never really memorized these rules, I have to think through the interactions.

**Other Coordinate Systems**

The Unity game engine is left-handed, OpenGL (and most math/physics tools) are right-handed. Why?

In a computer game, `x`

goes horizontal, `y`

goes vertical, and `z`

goes “into the screen”. This results in a left-handed system. (Try it: using your right hand, you can see `x`

cross `y`

should point out of the screen).

**Applications of the Cross Product**

- Find the direction perpendicular to two given vectors.
- Find the signed area spanned by two vectors.
- Determine if two vectors are orthogonal (checking for a dot product of 0 is likely faster though).
- “Multiply” two vectors when only perpendicular cross-terms make a contribution (such as finding torque).
- With the quaternions (4d complex numbers), the cross product performs the work of rotating one vector around another (another article in the works!).

Happy math.

## Other Posts In This Series

- Vector Calculus: Understanding the Dot Product
- Vector Calculus: Understanding the Cross Product
- Vector Calculus: Understanding Flux
- Vector Calculus: Understanding Divergence
- Vector Calculus: Understanding Circulation and Curl
- Vector Calculus: Understanding the Gradient
- Understanding Pythagorean Distance and the Gradient

TheophilusApril 15, 2015 at 10:35 amKalid has done it again! Imagine if we are taught like this in secondary schools, our university engineering education would have been so fantastic

Mistu sahaJuly 13, 2017 at 9:09 amSuperstition

kalidApril 15, 2015 at 10:50 amThanks Theophilus, glad you enjoyed it! I really wish someone showed me how the dot and cross product combine to build out the full picture.

MikeApril 15, 2015 at 2:39 pmYeah, very great explanatory series. Very helpful and intuitive. Unfortunately this is often not a norm even at universities. Thanks much ;)

JaeApril 15, 2015 at 5:54 pmI create two triangles with numbers at their edges, in order to represent two 3D vectors. I use them as base and top of a triangular cylinder. I draw two diagonal lines on each sides of the cylinder. Each of the 6 diagonal lines gets a number. Since I made my triangular representation of vector follow a counter-clockwise pattern, diagonal lines that go down, or approaches the base, as I go counter-clockwise gets (+) sign, and the diagonal lines that go up as I go counter-clockwise gets (-) sign. Each set of 3 pairs of diagonal lines crossing each other generates a number, giving us 3 pieces of information.

A perplexing point to me is that we choose to store the number we get by observing interaction of (y&z) and (z&y) into x! My childish guess is that x is neutral to y and z, so both y and z are happy to have their votes stored in x’s chest.

Haniff DinApril 19, 2015 at 5:38 pmActually this is the completely wrong way to teach this. In actual fact it’s actually properly taught by using Geometric Algebra. I.e clifford algebra.

Using the “proper way” leads to far more insight and intuition into how this is used and what it really means in actual use.

It’s a shame that it’s never taught properly.

Robert WilsonMarch 14, 2017 at 8:41 pmExcept that no textbook on linear algebra or vector calculus known to mankind has included clifford algebra when discussing cross products, normal vectors or determinants, and most of the people reading this site are going to be undergraduate students in these courses looking for clarity, help and explanations that look familiar enough to them to help them understand what their book is saying. To this end, I think Kalid succeeded dramatically.

kalidApril 21, 2015 at 9:20 am@Mike: Glad you enjoyed it!

@Jae: I might need to see a diagram, but yep — the idea is that x is perpendicular (not favoring) either y or z.

@Haniff: The idea is to show how the traditionally taught dot/cross product fit into a larger picture, but not to start with geometric algebra on day 1 (“I thought we were in a vector calculus class?”).

Mark PtakMay 17, 2015 at 8:24 amEver since my kids came home from school with the “lattice method” of multiplication and it took me a score of tries to “get it”, the true insight of using the distributive property to break numbers into smaller (or different) components has fascinated me. Kalid does it again with this nice graphic. Can’t wait for the day when he and Vihart are in the same real (or virtual) space.

kalidMay 17, 2015 at 3:35 pmThanks Mark, glad the matrix/lattice clicked. You might like this one too:

http://betterexplained.com/articles/how-to-understand-combinations-using-multiplication/

I’d love to collaborate with Vi one day.

gulrezMay 30, 2015 at 1:41 amKalid ,well done again, can we see result of complex multiplication in terms of total contribriution of components,I mean complex multiplication is multiplication of x and y coordinates, where similar coordinates multiplication indicates the total push (dot product) and different coordinates multiplication indicate magnitude (area,volume;cross product) and then we subtract the push which is useless( i square),plz comment ,thanks once again for insight

JacksonJune 3, 2015 at 3:07 pmGreat article but I’m a little confused on why the dot product is a number whereas the cross product is a vector. Could you elaborate further please? Also, off the topic, do you think you’ll ever do an abstract algebra section?

Thanks.

kalidJune 3, 2015 at 5:49 pmHi Jackson, great question. The dot product measures similarity (how similar is vector A to vector B?), and the result is a percentage. The assumption is you already have vector B, so a percentage is enough. If you like, you can do “(A dot B) times B” to get an actual vector.

The cross product asks “How different are A and B?”. Although the difference can be quantified as a percentage, there is ambiguity because two vectors can be entirely different from a given vector. (I.e., North and South are entirely different from East). Specifying the cross product as a vector means we can distinguish “The difference between North and East” from “The difference between South and East”.

Abstract algebra would a fun topic down the line, thanks for the suggestion.

JohnFebruary 2, 2018 at 3:45 pmHi,

Could you elaborate a bit more on the analogy where one vector is North and the other is East? Specifically, I’m asking about how specifying the cross product as a vector lets us distinguish “the difference between *North and East*” from “the difference between *East and North*”. I’m referring to how the cross product isn’t commutative.

Thanks.

tapan kumar dasJune 21, 2015 at 12:13 ammr. kalid simply you are a geneous of math. but on the contrary i am really a foolish one.

tapan kumar dasJune 21, 2015 at 12:15 ameinstein?

Ambrose wafula matsesheJuly 17, 2015 at 7:54 amPlease answer my question

furqanAugust 28, 2015 at 11:54 pmExcellent

JakeSeptember 12, 2015 at 2:49 amThis is a fantastic explanation and answer. I finally have a fundamental understanding and knowledge of the cross product after all these years. This is well-written and it all makes sense to me now. I will be coming back to this site again for more articles like these after what I’ve learnt today. Thank you.

kalidSeptember 15, 2015 at 11:28 pmThanks Jake, glad you enjoyed it. Hope you enjoy the rest of the site.

vivek kumar mauryaOctober 1, 2015 at 12:11 amw=fa why scalar quantities

SureshOctober 7, 2015 at 10:30 amYour articles are inevitably good brother!

They give a complete insight and provokes visualization of math equations.

Beyond everything, you are very good amazing person I feel.

Keep up the Best work;

kalidOctober 7, 2015 at 10:37 am@Suresh: Thanks, it’s awesome to hear when it helps. I’d like to help other people avoid the frustrations I had, appreciate the kind words.

AnonymousOctober 31, 2015 at 4:52 pmOh my gosh! Thank you so much! As an astronomy nut and physics student, I was terrified over my difficulty grasping the cross product. This is so much better than my professors and textbook

Ali mohammedNovember 30, 2015 at 10:24 amthank you for your explanation

but how we can find unit vector from a given points with out using cross product?

AnonymousNovember 30, 2015 at 10:57 amThank you kalid sir it is very useful for us because you explain very beautifully. I think it is amagine.. Again thank u so much.

AmeldaDecember 2, 2015 at 7:21 amThanks so much for this wonderfully lucid explanation! I have always been baffled by this.

John HallMay 18, 2016 at 7:20 amThis is missing a section on physical intuition.

kalidJune 3, 2016 at 1:46 pmGood question — the physical intuition is that the cross product tracks the area of the parallelogram spanned by two vectors. We get the most area when the vectors are perpendicular (i.e. the parallelogram is a square).

In a more abstract sense, the cross product is tracking the “non-similar interactions”, i.e. getting stronger the more different the vectors are. Sometimes we want to measure similarity (dot product), sometimes we want to measure differences (cross product).

AhmedFebruary 18, 2017 at 11:50 pmAll my respect to you sir

Let me share you my confusing that since we can measure the similarity (dot product) that also indicate the difference as well without (cross product),that is base on the idea of the similarity somehow indicate the differences.

sohaibJune 21, 2016 at 11:03 pmas u pointed out that in right hand rule there are two possible directions .i am bit confused in point that what is reason for chosing one direction not other.is it a random rule chosen by mathimatician or there is solid reason behind this plz .

PacoAugust 22, 2017 at 8:34 amIt is a random rule chosen by mathematicians. You could rewind history and choose it the other way around; the only effect would be to change millions of signs in all mathematics textbooks.

A very similar thing happens with the “magnetic field” in electromagnetism. Precisely because it is “defined” from a cross product and therefore inherits this arbitrariness. In physics, vectors that have this arbitrariness to them are called “pseudovectors”.

cnaJuly 30, 2016 at 12:48 amI just can say it was awesome! Felt transcending reading it,can’t imagine how you felt thinking it up and writing it!Kalid you are a genius! Please keep up the great work! Tnx

prenticeAugust 11, 2016 at 11:52 amHuh to ahaaaaa!actually

Xz12August 16, 2016 at 12:07 amI have a doubt . My book says that we can’t always calculate the angle between two vectors using vector or cross product.

But WHY ?

dl2020October 18, 2016 at 4:56 amAbsolutely brilliant. I was just getting confused with some matrix work and realized I didn’t really even understand what an inner product _was_. How can I be using all of these advanced formulas and not know the basics inside and out? Then I looked in Strang’s textbook for the answers and what I got was more of the same definitions and subtly contradictory examples with all the interesting detail glossed over… so I started working this stuff out myself and was just amazed at the beauty of what I was seeing and confused by the fact that I could not find it written out clearly anywhere!

Then I came across this – what a great job – this is true teaching, meeting the interested student on his own turf and starting from square one. If I had learned this way in school, I’d know 10x as much and really know it instead of just constantly looking up formulas. Keep up the good work.

KalidOctober 19, 2016 at 6:37 pmAwesome, really glad to hear it’s helping. Thanks for the note.

JacobOctober 28, 2016 at 6:52 amMultiplication with quaternions are associatieve. Maybe that is why they are easier to use in 3D space.

LucyOctober 31, 2016 at 7:03 amhi! May I know why cross product measures the “percentage difference”

LucyOctober 31, 2016 at 7:03 amhi! May I know why cross product measures the “percentage difference”?

nilsNovember 15, 2016 at 11:54 amThe explanations are wonderfully simple and easy. Congratulations. But, the color scheme on the entry pages need to change. Maroon and gray do not contrast enough. Suggest Maroon and white?

SughraNovember 26, 2016 at 11:58 pmWhy do we use sine theta in cross product??

Sarthak SharmaDecember 28, 2016 at 5:08 amHello Kalid , I enjoy learning from you much much better than anyone else , keep making such good videos , I posted a comment on your video , but you didn’t answer , so I am posting again here , in limit evaluation , the variable of limit is quite tricky , means does it need to be in correct place at correct time and if it is not then you get weird results ,see i was solving a definite integral and I was evaluating it as a Riemann sum(between limits 1 and 4) so it looked something like this =>

lim h->0 h [f(1)+f(1+h)+f(1+2h)+……….+f{1+(n-1)h} , now see if I say at this moment that “my limit says h is going to 0 , so due to this as h is outside this whole sum , then the whole sum should approach to 0 ?” but when i open the sum and do some algebra then some of the h cancel out and those who are left gives a non-zero result why this happens , why h needs to be stuck in a correct place before evaluating the limit.

Thankyou , I am hoping you will soon answer this

Bye!

Sarthak SharmaDecember 28, 2016 at 5:53 amHello again Kalid , I ran into a problem again , sorry to bother you but please tell me something about this. I was solving a cross product but soon i found out that when i evaluated it normally using basic vector algebra and when i did the same using a determinant , i got different results , determinant resulted in a whole shift in signs , i thought that the possible solution was that when i used the determinant those unit vectors were not the same as those in my normal vector equations , means somehow we need to take a whole opposite coordinate system relative to each other ( either relative to determinant or to my algebraic one ) , what do you think about this ?

Thank you.

purushFebruary 14, 2017 at 5:14 pmvery useful

marjathaleeFebruary 26, 2017 at 6:15 amYou made me fall in love with maths. Thank you

Robert WilsonMarch 14, 2017 at 8:33 pmThis is the most informative piece of work I have ever seen on the cross product and the theory of the cross product. I have learned quite a lot, and this has saved me hours of strife in trying to understand what the cross product represents and why it is the way it is. I feel much more comfortable using the cross product now that I understand it. Thanks!

Robert WilsonMarch 14, 2017 at 8:37 pmI’m curious to know, did you use Marsden and Tromba ch. 1 as your source for much of this stuff?

bilal shifawMarch 16, 2017 at 1:18 amephew!! i wish i saw this site earlier!! im grade 12 and i got this website very important to understand vector easly ways!! i am so greatfull to owner of this site!!

Transformation Fitting using the Horn method – NullbreakMarch 28, 2017 at 9:21 pm[…] closely related to the dot/cross product. For information about this, I’d recommend checking this very helpful website. As it turns out, the off-diagonal entries encode the cross product, while the diagonal entries […]

Representing plane intersections as a system of linear equations | Andy G's BlogApril 27, 2017 at 7:09 pm[…] It’s not the most intuitive computation. Personally, I enjoyed the “xyzxyz” explanation at Better Explained. […]

(:-)April 28, 2017 at 12:28 amWow!nice explanation….. Thank u so much,☺☺

ErsinMay 14, 2017 at 6:14 amThese explanations are perfect for the lazy high schooler like me that loves understanding concepts and exploring but hates to meaninglessly memorize formulas, I love your intuitive explanations and creative presentations!

Sara AsfarMay 29, 2017 at 8:48 pmFor which angle cross product is zero;0 or 180?

Sara AsfarMay 29, 2017 at 8:50 pmWhat is the right answer and what is reason behind it?

AnonymousJune 12, 2017 at 12:26 amAnother simple application you could add is that when the cross product is zero this indicates parallel vectors which fits with your intuitive “cross interactions” idea.

Issa khreisJune 29, 2017 at 3:36 amHow do we verify that a point doesnt belong to a plane

hiddenbananaJuly 18, 2017 at 12:54 pm“Whoa! How’d we get back to j? ” I would like to now that, since I thought k x i = -j

kalidJuly 27, 2017 at 9:52 pmBy the “xyzxyz” order, zx is y (positive).

Try it on Wolfram Alpha here:

http://www.wolframalpha.com/input/?i=(0,+0,+1)+cross+(1,+0,+0)

WALI MOHAMMADJuly 22, 2017 at 7:04 amVery nicely explained, though i could not understood the whole concept but still will help me a lot.

kennethoneAugust 9, 2017 at 1:56 pmYou forget that a cross product between two vectors describes the equation of a line going through the ends of those vectors. See: https://math.stackexchange.com/questions/1973353/how-to-calculate-the-parameters-of-a-line-through-two-points-with-homogeneous-co

150 多个 ML、NLP 和 Python 相关的教程 | Hello word !August 21, 2017 at 5:47 pm[…] Understanding the Cross Product (betterexplained.com) […]

PacoAugust 22, 2017 at 8:29 amVery nice!

Suggestion: in the “Dot Product of Cross Products” section you can explain why a.(bxc) represents the volume of the parallelepiped created by a, b and c.

In turn this gives intuition into the “commutation” rule:

a.(bxc) = c.(axb) = b.(cxa)

because they all describe the same volume.

This commutation relation simplified enormously a problem that I was having in a serious research in nanophotonics. And I did not know about it. I only found it coincidentally.

TwenteMasterSeptember 4, 2017 at 12:43 amAfter read this article 3 times, I still can not understand cross product. I think this one is not as good as the “understanding the dot product” one.

when I read the “Defining the Cross Product” section, I were stucked. I can’t even know what does this section mean? so I couldn’t continue to understand the next several sections.

Kalid, can you please rewrite this article in a more simple way? just as you did in the “understanding the dot product”.

I also want to know that is there only I can’t understand this article.

Thanks!

ScienceDiscovererOctober 25, 2017 at 9:58 amI made a slightly better version of your 3×3 diagram! It shows that one side of it is negative, and other is positive. Very intuitive! https://i.imgur.com/BnGScqC.jpg

ScienceDiscovererOctober 25, 2017 at 9:59 amAlso, the same color squares represent what parts of vectors are “voting” for what part of cross product vector!

GaharietNovember 14, 2017 at 11:13 amI don’t understand your computation of cross product (!). Cross product between two vectors should return a matrix (computed from its outer products)! But instead we get a new vector…

KalidNovember 14, 2017 at 10:02 pmWhile the cross product can be seen as having 6 components from the matrix, the cross product is defined as a vector. Similarly, the dot product is defined as a scalar (even though it is made from 3 terms).

KendraDecember 16, 2017 at 2:06 amI needed to learn how to manually do a cross product for an exam tomorrow. This made it possible. Thank you!!!!

calFebruary 13, 2018 at 1:07 pmHow would I calculate |u x v| ? Assuming u is (u1, u2, u3) and v is (v1, v2, v3)? Trying to calculate this equation:

u x v / |u x v|

Been banging my head against a wall on this! Any help you can offer would be greatly appreciated!

kalidFebruary 13, 2018 at 3:20 pmHi Cal,

The goal of u x v/|u x v|

is to find the unit normal vector (the full cross product, scaled down to a length of 1).

The cross product has length |u||v|sin(theta), which we can find by directly crunching through the cross product formula.

As a shortcut,

Because cos^2 + sin^2 = 1, we have

sin = sqrt(1 – cos^2)

From the dot product,

u . v = |u||v|cos(theta)

or

cos(theta) = u.v / (|u||v|)

Hope that helps!

Laurence QiJuly 24, 2018 at 11:52 pmHonestly this is such a phenomenal explaination of a topic that I’ve had significant difficulty with in the past

Humoyun AhmedovNovember 27, 2018 at 11:36 pmthis one is not intuitive, more complicated than other posts on this series