I know the question seems kind of unclear but why (in the attached image) the cross product like this, I hope someone can clear my doubt.
122731
The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:  $$\left| \begin{matrix} \vec{i} &amp; \vec{j} &amp; \vec{k} \\ a_i &amp; a_j &amp; a_k \\ b_i &amp; b_j &amp; b_k \end{matrix}\right|$$  Is there any physical significance to this matrix, or is it just some mathematical trick?
54080
I do not know an example. Will ask question if in doubt of the proofs provided thank you!!
25156
I haven't really formally studied Algebra at anywhere near this level, but I was told about the existence of Quaternions a few years ago and I find them really cool. I also like how pure quaternions are analogous to cross products in $R^3$, and that it gives me a way of doing it algebraically rather than relying on hand rules.  Over the course of using them it's my understanding that multiplying an element by another, both imaginary numbers, rotates it in some manner so as to be orthogonal to both, similar to rotations of $\frac{\pi}{2}$ in the space with one real and one imaginary axis when multiplied by $i$, which is why it's useful for cross products as it will rotate it to be orthogonal to a plane spanned by linear combinations of those two vectors [afaik].   What I can't intuit though is why for any pure quaternion multiplication, there's a negative real part [the scalar product] if it's meant to correspond to a rotation within $R^3$ and the space of pure Quaternions is Isomorphic to $R^3$. Similarly I don't understand why $i^2 = j^2 = k^2 = $ A negative real.
64768
I think it has to do with $\sin$ and $\cos$ because most of them have three limits but I couldn't find a good example that it is wrong although I think that it is indeed wrong.
21573
I am learning linear algebra, and I am a bit confused by the dot product and how the answer to the process turns out to be a scalar rather than a matrix.  For $2$ vectors with $2$ components, I learned that dot product is equivalent to a $1 \times 2$ row vector left multiplied by a $2 \times 1$ column vector.   The result of such a multiplication should result in a $ 1 \times 1 $ vector.   But I am learning that the dot product somehow transforms the result into a scalar, rather than a $ 1 \times 1 $ vector.  Maybe I am missing something but the difference between a $ 1 \times 1 $ vector and a scalar seems important, because you can multiply a scalar by a matrix of any size, but you can only left-multiply a $ 1 \times 1 $ vector by another matrix with $1$ row.  Thanks for any help in understanding this.
27702
Wikipedia introduces the vector product for two vectors $\vec a$ and  $\vec b$ as $$ \vec a \times\vec b=(\| \vec a\| \|\vec b\|\sin\Theta)\vec n $$ It then mentions that $\vec n$ is the vector normal to the plane made by $\vec a$ and  $\vec b$, implying that $\vec a$ and  $\vec b$ are 3D vectors. Wikipedia mentions something about a 7D cross product, but I'm not going to pretend I understand that.  My idea, which remains unconfirmed with any source, is that a cross product can be thought of a vector which is orthogonal to all vectors which you are crossing.  If, and that's a big IF, this is right over all dimensions, we know that for a set of $n-1$ $n$-dimensional vectors, there exists a vector which is orthogonal to all of them.  The magnitude would have something to do with the area/volume/hypervolume/etc. made by the vectors we are crossing.  Am I right to guess that this multidimensional aspect of cross vectors exists or is that last part utter rubbish?
41230
I've just started studying linear algebra, and I'm having trouble understanding something that apparentely seems obvious to everyone else. My question is, why does multiplying a vector by a positive scalar does not change its direction? I've seen some visual representations of this and it does seem reasonable to me that that should be the case. However, I wonder whether there is a way to actually show it mathematically? In the 2d coordinate plane, for instance, I guess way one of showing that vector $A$ has the same direction of $cA,c&gt;0$ would be to show that the line containing $A$ has the same slope as the one containing $cA$ (I'm not sure if this approach is correct though). However, how to proceed in the case of $n$ dimensions? I think I must be missing out something basic here, but I really can't seem to get my head around it. Any help will be highly appreciated. Thanks very much in advance.
16281
Most questions usually just relate to what these can be used for, that's fairly obvious to me since I've been programming 3D games/simulations for a while, but I've never really understood the inner workings of them... I could get the cross product equation as a determinant of a carefully-constructed matrix,   but what I want to ask is... How did the dot and cross product come to be? When were they "invented"? Some detailed proofs? Did someone say: "Hey, wouldn't it be nice if we could construct a way to calculate a vector that is perpendicular to two given operands?"  Basically, how/why do they work?  I would appreciate explanations, links to other explanations, other web resources... I've been searching the Internet lately for explanations, but most of them are on how to use it and nothing that really gives substance to it.
54519
I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.  A non-graphical calculation will be very helpful too.  Thank you so much in advance!
54549
Hello,  in the picture there is an equation I can not understand. What I do not understand is how to get to the first step. I have been thinking and thinking but can not find an explanation. Could somebody please explain? Thanks.
6953
Could someone please provide an explanation for the origin of ? (Ideally someone who I have been speaking with on the h-bar)  Any advanced maths beyond basic calculus will most probably leave me at a loss, though I do not mind a challenge! Please assume little prior knowledge, as over the past few days I have discovered that much of my understanding surrounding the process as virtual particle pairs is completely wrong.
71750
Note that a product   may be parenthesized in two different ways:   and . Similarly, there are several different ways to parenthesize . Two such ways are  and . Let  be the number of different ways to parenthesize the product . Show that if , then      for all integers .  Keep in mind that the sequence  is the same as the sequence of Catalan numbers.   I am entirely lost. Please can someone offer some hint or direction of where to begin? I humbly thank you so very much.
45671
Hi I'm having trouble understanding why the quaternionic group cannot be written as a semi-direct product in a non-trivial way. Is there a proof of why this is the case?
36899
how is this possible? I know there is some trick, should someone please explain?!
87286
Just reading about the cross product. It says here: $$A\times B= \{(x,y)|x\in A, y\in B\}$$ Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?    Edit: Quoting the free textbook I'm looking at online,     If we are given that A is a set and no other information about A, then there is no   ordering to the elements of A. Thus, we cannot speak of “the second element of the set A”   unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.       Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated,   but it is not okay to have repeated elements in an ordering of S?   When we say S  =   {a, b, c, a}, we know that S contains just the three elements a, b and c.  If we were to talk   about the ordered set (a, b, c, a) it would not make sense because it would say that the   element a is in two places at once: the first position and the last position.
97089
We know how to make cross product of three dimensional vectors. $$ \vec A \times \vec B = \vec C$$  Where : $ \vec A = (A_i; A_j; A_k)$ $ \vec B = (B_i; B_j; B_k)$ $ \vec C = (C_i; C_j; C_k)$  $C_i = \left|\begin{matrix}A_j&amp;A_k\\B_j&amp;B_k\end{matrix}\right|$ $C_j = \left|\begin{matrix}A_k&amp;A_i\\B_k&amp;B_i\end{matrix}\right|$ $C_k = \left|\begin{matrix}A_i&amp;A_j\\B_i&amp;B_j\end{matrix}\right|$  But what about if we have four dimensional vectors? Is it possible to make cross product of four dimensional vectors? If it  is possible, then tell me when it can be possible?       Let say we have two vectors: $ \vec A = (A_i; A_j; A_k; A_l)$ $ \vec B = (B_i; B_j; B_k; B_l)$   Then how to compute a cross product of this two vectors? Will it again vector? $$ \vec A \times \vec B = \vec C$$  $ \vec C = (C_i; C_j; C_k; C_l)$ Then how to compute those coordinates?  We know that only square matrices have a determinant property! In this case it might not be correct if we will wright... $\color{red} {\text { $C_i = \left|\begin{matrix}A_j&amp;A_k&amp;A_l\\B_j&amp;B_k&amp;B_l\end{matrix}\right|$}    C_j = \left|\begin{matrix}A_k&amp;A_i&amp;A_l\\B_k&amp;B_i&amp;B_l\end{matrix}\right| C_k = \left|\begin{matrix}A_i&amp;A_j&amp;A_l\\B_i&amp;B_j&amp;B_l\end{matrix}\right| C_l = \left|\begin{matrix}A_i&amp;A_j&amp;A_k\\B_i&amp;B_j&amp;B_k\end{matrix}\right|}$   So tell me how to solve this problem?
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.