# A Bunch of Questions about Vectors **URL:** \ **Category:** Factual Questions\ **Created:** [August 26, 2002, 9:08pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680 "2002-08-26T21:08:16Z")\ **Posts on this page:** 13\ **Page:** 2
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [August 27, 2002, 9:25pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/21 "2002-08-27T21:25:52Z")
Ok, this has been helpful, but I’ve hit another snag. Im trying to get: (-8 **i** + 5 **j** - 4 **k** ) dot (3 **i** + 3 **j** -2 **k** ) Should I distribute this? That seems to be what my book wants. But then I get for the first part of the distribution: (-8 **i** dot 3 **i** ). According to the book, I should do (-8)(3)(cos theta.) Since they’re in the same direction, the cosine (of zero) should be 1, leaving me with -24. But in what direction? I don’t quite see how the method **ultrafilter** gave could be used here. Also, is there a way to do dot products and cross products on a TI-89 calculator? I obviously will still learn how to do them by hand, but I think that would be good for checking my answers and for taking the actual AP test. Thanks again for all the help. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 27, 2002, 9:26pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/22 "2002-08-27T21:26:54Z")
The dot product of two vectors has no direction, as it’s a scalar. ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [August 27, 2002, 9:35pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/23 "2002-08-27T21:35:31Z")
**Qwerty** , dotproducts don’t have a direction; they’re what is known as a _scalar_, i.e. just a number. Crossproducts are vectors (really, pseudo-vectors, but that’s not important at the moment), and do. In order to program dot products, the easiest thing to do is to make a pair of 3x1 matrices and put the components of each vector in one of the matrices… so if you had (x,y,z) and (u,v,w), you’d define matrices A=(x,y,z) and B=(u,v,w). Then take A\*B[sup]T[/sup] in your calculator and you’ll get the dotproduct. For the crossproduct, it’s best just to program the formula **zut** gave. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 27, 2002, 9:35pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/24 "2002-08-27T21:35:34Z")
Let me run through an example, that might make it a bit easier. I’m gonna use a . b for a dot b. First, there’s an error in what you’re doing. 3i and -8i are not in the same direction; they’re 180 degrees apart. The formula in your book is a . b = |a||b|cos(t), where |a| is the legnth of a, and t is the angle between a and b. There are two ways to handle (-8i + 5j - 4k) . (3i + 3j -2k). You can distribute out, or you can evaluate the vectors and do it that way. (-8i + 5j - 4k) = -8(1, 0, 0) + 5(0, 1, 0) - 4(0, 0, 1) = (-8, 5, -4). Similarly, (3i + 3j -2k) = (3, 3, -2). So (-8i + 5j - 4k) . (3i + 3j -2k) = (-8, 5, -4) . (3, 3, -2), which is equal to (-8 \* 3) + (5 \* 3) + (-4 \* -2). That evaluates to -1, unless I’ve made an arithmetic mistake. If you were to distribute out, you’d discover that ai . bj = 0, bj . ck = 0, and ck . ai = 0 (remember that the dot product is commutative, so that’s everything). So you get (-8i . 3i) + (5j . 3j) + (-4k . -2k). Since i . i = j . j = k . k = 1, it works out to be -1 again. Does this make sense? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 27, 2002, 9:36pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/25 "2002-08-27T21:36:39Z")
> [@](#): > > \*Originally posted by g8rguy \* > **Crossproducts are vectors (really, pseudo-vectors, but that’s not important at the moment), and do.** Pseudovectors? ---
**Author:** ![Qwertyasdfg](https://avatars.discourse-cdn.com/v4/letter/q/ecc23a/32.png) [@Qwertyasdfg](https://boards.straightdope.com/u/Qwertyasdfg)\ **Post date:** [August 27, 2002, 9:48pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/26 "2002-08-27T21:48:58Z")
Ok **ultrafilter** that really helped to clear things up. Thanks again everyone. ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [August 27, 2002, 10:41pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/27 "2002-08-27T22:41:41Z")
Ah, sorry, that’s a physics term. Basically, it has to do with the way it transforms under **x** -\> - **x** , which is important in categorizing things in physics. Since a cross product doesn’t pick up a minus sign under this transformation, i.e. if **a** -\> - **a** and **b** -\> - **b** , then **a** x **b** = **c** -\> (- **a** ) x (- **b** )= **c** , it transforms differently than a regular vector, so we call it a pseudo-vector. (You’ll also hear axial vectors and polar vectors, but I never remember which is which.) ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 27, 2002, 10:44pm UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/28 "2002-08-27T22:44:08Z")
Gotcha. Thanks. ---
**Author:** ![JS\_Princeton](https://avatars.discourse-cdn.com/v4/letter/j/57b2e6/32.png) [@JS\_Princeton](https://boards.straightdope.com/u/JS_Princeton)\ **Post date:** [August 28, 2002, 1:37am UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/29 "2002-08-28T01:37:11Z")
> [@](#): > > \*Originally posted by ultrafilter \* > \*\* > > I’ve done a bit of work with Cartesian products, and I don’t see how they’re a generalization of cross products. Could you explain? \*\* Well, I’m a bit rust on my linear algebra, but we’ll see if I can spit out what I’m trying to say… IIRC, if you have an _n_-dimensional vector space and two vectors that are linearly independent, then you can define an operation similar but more general than the cross product using Cartesian products. My linear algebra vocab is rusty, but I believe what you will find is that these Cartesian products will always be a third vector in the nullspace of the the first two vectors. Of course, I may be barking up the wrong tree here, and my linear algebra text has decided to migrate from my bookshelf… still I am confident I learned about the generalization of the cross product at some point. I was sort of hoping some mathematician would jump on board and really give me a good reminder of what this is all about, but I’m afraid I may have to use printed resources instead. SIGH… ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 28, 2002, 3:53am UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/30 "2002-08-28T03:53:03Z")
> [@](#): > > \*Originally posted by JS Princeton \* > \*\* > > Well, I’m a bit rust on my linear algebra, but we’ll see if I can spit out what I’m trying to say… > > IIRC, if you have an _n_-dimensional vector space and two vectors that are linearly independent, then you can define an operation similar but more general than the cross product using Cartesian products. My linear algebra vocab is rusty, but I believe what you will find is that these Cartesian products will always be a third vector in the nullspace of the the first two vectors. Of course, I may be barking up the wrong tree here, and my linear algebra text has decided to migrate from my bookshelf… still I am confident I learned about the generalization of the cross product at some point. I was sort of hoping some mathematician would jump on board and really give me a good reminder of what this is all about, but I’m afraid I may have to use printed resources instead. SIGH… \*\* I wouldn’t be surprised if such a thing showed up in tensor theory, one of those many areas that I know nothing about. We’ll see if someone knows it, or if you can find a reference, that’d be cool. ---
**Author:** ![Joe\_Cool](https://avatars.discourse-cdn.com/v4/letter/j/ecb155/32.png) [@Joe\_Cool](https://boards.straightdope.com/u/Joe_Cool)\ **Post date:** [August 29, 2002, 3:40am UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/31 "2002-08-29T03:40:18Z")
Ok, I have a couple questions. I dived back into my linear algebra book (Cliff’s Notes. Don’t laugh. I’m good enough at math that I think I can get it with just an outline, normally), and I’m up to dot-products and cross-products. Now… The cross-product: My book says it can only be done in R[sup]3[/sup] space. Is that true, or is the R[sup]3[/sup] cross-product a subset of some larger generalized thing? And the dot-product: what is its geometric interpretation? My book says it’s related to the cosine through something called the Cauchy-Schwartz inequality: -1 \< ([sup]x . y[/sup]/[sub]||x|| ||y||[/sub]) \< 1). Now, is that the only direct geometric relation to the vectors x and y, or is there something else? AND, why is it related to the cosine? How is it derived? I assume it comes from the fact that if x and y are perpendicular then x . y is 0, but how do you proceed from there? Sorry to be asking such noob questions. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 3:50am UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/32 "2002-08-29T03:50:26Z")
x . y = |x||y|cos([symbol]q[/symbol]) ---
**Author:** ![JS\_Princeton](https://avatars.discourse-cdn.com/v4/letter/j/57b2e6/32.png) [@JS\_Princeton](https://boards.straightdope.com/u/JS_Princeton)\ **Post date:** [August 29, 2002, 4:26am UTC](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680/33 "2002-08-29T04:26:17Z")
> [@](#): > > The cross-product: My book says it can only be done in R3 space. Is that true, or is the R3 cross-product a subset of some larger generalized thing? Once I track down MY linalg text I’ll be able to tell you in greater detail what the cross product is a subset of. Please see **ultrafilter** ’s and my discussion on the topic above. > [@](#): > > And the dot-product: what is its geometric interpretation? Think of two vectors that have an angle between them. If you remember from geometery the law of cosines, you can actually use vector rules and dot-products to derive it. The geometric interpretation is just that… sort of the left over bit that comes into play when trying to play Pythagorean Rules with angles that are not ninety degrees. Another way to say this is that the dot product is the length of a projection of either vector onto the unit vector of the other. That’s why it’s related to a cosine because when the two vectors are perpendicular there is no projection. [Here’s a site](http://mathworld.wolfram.com/DotProduct.html) with all the salient proofs for dot products. The other way to multiply vectors is through cross products. This defines a third vector which goes in the direction of the right hand rule, perpendicular to both of the vectors in question (normal to the plane they define since two vectors determine a plane), and the resultant vector is defined to have length equal to the area of the parallelogram defined by the two vectors you are multiplying. [Previous page](https://boards.straightdope.com/t/a-bunch-of-questions-about-vectors/125680.md?page=1)