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We have to pick one way to do it - either the right hand rule or the left hand rule. It's not important which one we use, as long as everyone uses the same one - just like which side of the road we drive on. We'd get the same eventual results if we all used the left hand rule. Someone chose the right hand rule over the left hand rule, and it became ...


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It is the way the cross product and the convention of the coordinate system is defined. https://en.wikipedia.org/wiki/Right-hand_rule#Coordinates


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No to both questions. It's usually easier to use the dot product $A\cdot B = |A|\,|B|\,\cos\theta$. It is only equivalent if either $\sin\theta=0$ or $A\cdot B=|A|\,|B|$ which happens when $\cos\theta=1$. (So, almost never..)


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Using the distributive law, $$(A+B)\times(A-B)=A\times A-A\times B+B\times A-B\times B$$ $$A\times A=B\times B=0$$ $$(A+B)\times(A-B)=0-A\times B+B\times A-0$$ $$B\times A=-A\times B$$ $$(A+B)\times(A-B)=-A\times B-A\times B=-2(A\times B)$$


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It's better if you define $F$ in terms of smooth functions in each coordinate. For instance I would write $F = (F_x, F_y, F_z) = F_x\hat{i} + F_y \hat{j} + F_z \hat{k}$ and compute each quantity one at a time. First you'll compute the curl: $$ \nabla \times F \;\; =\;\; \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & ...


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Since you've included the “cross-product" tag, here's a way to do this with cross products: Sort the vertices of the triangle in order of increasing angle. Slope will do since all of the points are in the first quadrant. Call these points, in order, $P_1$, $P_2$ and $P_3$. Side $P_1P_3$ is the only one visible if it is closer to the observer than $P_2$, ...


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For each of the vertices $V_i$, find the angle $OV_i$ makes with the $x$-axis. Compare them and you should be able to find the vertices that make the largest and the smallest angles respectively. From the vertex that makes the largest angle, counting anti-clockwise the edges of the triangle until the vertex that makes the smallest angle, and these edge(s) ...


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Because you can always make $v = y \times a/|a|$ and then $a \times v /|a|$ will give $y$, so $x = v/|a| = y \times a/|a|^2$


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Write $ y = y_1e_1+y_2e_2 + y_3e_3 $ and $ a = a_1e_1 + a_2e_2 + a_3e_3 $, where $\{e_1,e_2,e_3\}$ is the canonical basis for $\mathbb{R}^3$. Consider the matrix $$ A = \begin{pmatrix} 0 & a_3 & -a_2\\ -a_3 & 0 & a_1\\ a_2 & -a_1 & 0 \end{pmatrix} $$ and prove that the any non-trivial solution of $Ax = y$ is the $x$ you are looking ...


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First of all $(\nabla \times \vec{v})_i = \epsilon_{ijk}\partial_j v_k$ with Einstein summation. Then for any scalar $\phi$ $$ (\nabla \times \phi\vec{v})_i = \epsilon_{ijk}\partial_j (\phi v_k)= \epsilon_{ijk}(\partial_j \phi) v_k + \phi\epsilon_{ijk} \partial_j v_k= (\nabla\phi \times \vec{v})_i + \phi (\nabla\times \vec{v})_i $$ the first one is because ...


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$n \perp P$ and $v \parallel P$, so $n \perp v$. Then $L \in P$ and $n \perp P$, so $n \perp L$ and $v_1 \parallel L$, so $n \perp v_1$. So $\DeclareMathOperator{span}{span}n \in \span(v, v_1)^\top$. Further $v\times v_1 \perp v$ and $v \times v_1 \perp v_1$. This is a property of the vector product. So $v \times v_1 \in \span(v, v_1)^\top$. ...


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@MvG has answered it and I accepted it. I just wanted to post a picture I've drawn for this to help me understand in case if someone else finds it useful.


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Consider three-dimensional space $\mathbb R^3$. If you normally homogenize by appending $z=1$, that means that your geometry as you know it happens on the $z=1$ plane in space. But it's also possible to view geometric elements as linear (i.e. containing the origin) subspaces of the whole three-dimensional space. A point on the plane corresponds to a line ...


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You can also use the Gram-Schimdt process. Knowing $w$ and starting with a random vector $u$, you can find $x$ by writing $x = u - (w^T*u)/(w^T*w)*w$. Matlab example: >> w = rand(4,1) w = 0.9575 0.9649 0.1576 0.9706 >> u = rand(4,1) u = 0.4218 0.9157 0.7922 0.9595 >> x = u - (w'*u)/(w'*w)*w x = -0.3755 ...


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Assuming $w_1$ is non-zero: For any other $w_j$ non-zero you will have a solution with $x_1=1$, $x_j=-\dfrac{w_1}{w_j}$ and other $x_k=0$. You can then take combinations of such solutions to produce even more solutions. If any of the $w_j$ are zero, you can also add to these solutions vectors where the corresponding $x_j$s take any values. So in ...


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Because you have a underdetermined equation (less equations than unknowns) you need to assign dummy variables. lets say our vecotr $w=(w_1,w_2,w_3,w_4)^T$. Then $w^Tx=w_1+w_2x_2+w_3x_3+w_4x_4=0$. We know that we have one equation and 3 unknows. Hence, we can choose 3-1=2 free variables. Let us call $x_2=a$ and $x_3=b$. Now solve the equation for $x_4$. ...


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Note that $$\frac{\sqrt{3}}{2}\left(\vec{b}-\vec{c}\right)=\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}.$$ Since $\vec{b}$ and $\vec{c}$ are not parallel, $\vec{a}\cdot\vec{b} = -\frac{\sqrt{3}}{2}$, which implies that $\|\vec{a}\|\|\vec{b}\|\cos\theta = -\frac{\sqrt{3}}{2}$. Since $\vec{a}$ and ...


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Think thats IIT question hint $a\times b\times c=(a.c)b-(a.b)c$ can you solve now?


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Let $i_x, i_y, i_z$ be the basis vectors of a right-handed coordinate system. Now we can write \begin{equation} i_x \times i_y = i_z \quad i_y \times i_z = i_x \quad i_z \times i_x = i_y \end{equation} By the definition of vector product if the vectors are orthogonal the $sin (\theta) = 1$ while if the vectors are parallel $sin (\theta) = 0$ then ...


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First notice that if we define $(u \wedge v ) \cdot w = \det (u,v,w)$ and let $i,j,k,l = 1,2,3$ then $$(e_i \wedge e_j) \cdot (e_k \wedge e_l) = \begin{vmatrix}e_i \cdot e_k & e_j \cdot e_k \\e_i \cdot e_l & e_j \cdot e_l\end{vmatrix}$$ Consequently $$(u \wedge v)\cdot (u \wedge v)=|u \wedge v |^2 = \begin{vmatrix}u \cdot u & v \cdot u \\ u ...



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