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Can someone help me understand the cross product a little better, for me it makes more sense for the new vector to be somewhere between the original vectors and closer to the bigger one but that would be just adding to one another right? how does crossing say for example vector $\vec A \times \vec B =\begin{bmatrix} 1 \\ 0 \\ 0 \\\end{bmatrix} \times \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\\end{bmatrix}= \vec C$?


marked as duplicate by David K, user137731, user147263, Andrew D. Hwang, Najib Idrissi Oct 16 '15 at 16:40

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    $\begingroup$ What definition of the cross product did you learn? Are you having difficulty seeing how the definition leads to a perpendicular vector, or do you think "cross product" should have been defined differently? $\endgroup$ – David K Oct 16 '15 at 12:59
  • $\begingroup$ I don't understand how multiplying different dimensions which is the crossing part of the cross product if I understand correctly leads to another dimension? so I guess how or where does the perpendicular vector come from and why wouldn't it be in the same plane as the original vectors? $\endgroup$ – Raed Oct 16 '15 at 13:04
  • $\begingroup$ I don't know what you mean by "multiplying different dimensions" -- sounds pretty sci-fi, though. 😉 Maybe you should check out my answer to this question. $\endgroup$ – user137731 Oct 16 '15 at 13:14
  • $\begingroup$ @Raed Oh. You're the guy who asked this question yesterday. Seriously spend some time reading through my (way too long) answer to the question I linked to. It might help you with some of your questions about the dot and cross products. $\endgroup$ – user137731 Oct 16 '15 at 13:25
  • $\begingroup$ There are two well-known ways to define the cross product. One way is to say that the cross product of two vectors is perpendicular to them by definition. The other way is to give a formula; if you use that formula to compute $\langle 1,0,0 \rangle \times \langle 0,1,0 \rangle$ you can confirm that indeed the answer is \langle 0,0,1 \rangle$. We already have many questions with many answers about these definitions; if there is something missing from those answers, please be more specific about what it is. We don't want to just be repeating the answers that were already written. $\endgroup$ – David K Oct 16 '15 at 13:30

The cross product can be defined using a determinant $$ \vec{A}\times\vec{B} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat {k} \\ A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3\end{array}\right| $$ So, when dot this result with $\vec{C}$ you get the same as replacing $\hat{i}$ with $C_1$, etc. That means, $$ (\vec{A}\times\vec{B})\cdot\vec{C}=\mbox{det}(\vec{C},\vec{A},\vec{B}), $$ where the notation on the right is the determinant whose first row is $\vec{C}$, whose second row is $\vec{A}$ and whose third row is $\vec{B}$. And you know that determinants of matrices with two identical rows are $0$. That gives you $$ (\vec{A}\times\vec{B})\cdot \vec{A}=0\\ (\vec{A}\times\vec{B})\cdot \vec{B}=0. $$ In other words, the cross product is orthogonal to the original two vectors.


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