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I'm writing some notes for a honors physics class and I am having some trouble with some proofs. Say $\vec{A}$ and $\vec{B}$ are some geometric vectors. Then we defined the dot product $\vec{A}\cdot\vec{B}$ as the product of the proyection of $\vec{A}$ onto $\vec{B}$ with the magnitude of $\vec{B}$, and the cross product $\vec{A}\times\vec{B}$ as the vector whose length is the area of the parallelogram defined by $\vec{A}$ and $\vec{B}$ and whose direction was defined through the right hand rule. Then it was easy to show that $\vec{A}\cdot\vec{B}=\|\vec{A}\|\|\vec{B}\|\cos{\theta}$ and $\|\vec{A}\times\vec{B}\|=\|\vec{A}\|\|\vec{B}\|\sin{\theta}$. Non the less, when going to a cartesian coordinate system and writing $$\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$$ and $$\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$$I came into some trouble. When trying to show the formula for the dot product I used the argument $$ \begin{align} \vec{A}\cdot\vec{B} &=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\cdot(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})\\ &=A_x\hat{i}\cdot B_x\hat{i}+A_x\hat{i}\cdot B_y\hat{j}+A_x\hat{i}\cdot B_z\hat{k}+A_y \hat{j}\cdot B_x\hat{i}+A_y\hat{j}\cdot B_y\hat{j}+A_y\hat{j}\cdot B_z\hat{k}+A_z\hat{k}\cdot B_x\hat{i}+A_z\hat{k}\cdot B_y\hat{j}+A_z\hat{k}\cdot B_z\hat{k}\\ &=A_x\hat{i}\cdot B_x\hat{i}+A_y\hat{j}\cdot B_y\hat{j}+A_z\hat{k}\cdot B_z\hat{k}\\ &=A_xB_x+A_yB_y+A_zB_z \end{align}$$ But I wasn't able to demonstrate from the geometric definition that the dot product obeys the product rule I used in the first step like normal products do (I think I just need to show that the dot product is distributive under addition, I just don't know how). Likewise, I haven't found a convincing argument for the formula $$\vec{A}\times\vec{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \\ \end{vmatrix} $$ If anybody could help me prove the formula above and the distribution law under addition for dot products that would be great! (Side question: how do you write in latex cartesian unit vectors such as $\hat{i}$ so that the dot above the $i$ doesn't appear)

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  • $\begingroup$ Search Khan Academy for some videos which explain these results. $\endgroup$ – Simon S Apr 18 '15 at 22:46
  • $\begingroup$ For the latex question use \hat{\imath} and \hat{\jmath} to get $\hat{\imath}$ and $\hat{\jmath}$. $\endgroup$ – Joelafrite Apr 18 '15 at 22:52
  • $\begingroup$ Here's the cross product video, linking the determinant to the geometric definition on Khan. khanacademy.org/math/linear-algebra/vectors_and_spaces/… $\endgroup$ – zahbaz Apr 18 '15 at 23:17
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Just use the same approach as for the dot product, but let $\hat{i}\times\hat{i}=0$, etc. Let $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, and $\hat{k}\times\hat{i}=\hat{j}$. Finally let $\hat{i}\times\hat{j}=-\hat{j}\times\hat{i}$ etc. $$ \begin{align} \vec{A}\times\vec{B} &=(A_x\hat{i}+A_y\hat{j}+A_z\hat{k})\times(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})\\ &=A_x\hat{i}\times B_x\hat{i}+A_x\hat{i}\times B_y\hat{j}+A_x\hat{i}\times B_z\hat{k}+A_y \hat{j}\times B_x\hat{i}+A_y\hat{j}\times B_y\hat{j}+A_y\hat{j}\times B_z\hat{k}+A_z\hat{k}\times B_x\hat{i}+A_z\hat{k}\times B_y\hat{j}+A_z\hat{k}\times B_z\hat{k}\\ &=0+A_x B_y\hat{k} + -A_x B_z\hat{j}+\\ &\;\;-A_y B_x\hat{k} + 0 + A_y B_z\hat{i}+\\ &\;\;\;\;\;\;A_z B_x\hat{j} + -A_z B_y\hat{i}+0\\ &=(A_y B_z -A_z B_y )\hat{i} + (A_z B_x-A_x B_z)\hat{j} + (A_x B_y-A_y B_x)\hat{k} \end{align} $$

Which is equal to the determinant that you show.

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Actually the answer is here http://math.oregonstate.edu/bridge/papers/dot+cross.pdf Figures 3 and 8 are the key!!!!!

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