# Tag Info

2

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..)

0

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: $$\nabla \times F \;\; =\;\; \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & ... 1 Since you've included the “cross-product" tag, here's a way to do this with cross products: Sort the vertices of the triangle in order of increasing angle. Slope will do since all of the points are in the first quadrant. Call these points, in order, P_1, P_2 and P_3. Side P_1P_3 is the only one visible if it is closer to the observer than P_2, ... 1 For each of the vertices V_i, find the angle OV_i makes with the x-axis. Compare them and you should be able to find the vertices that make the largest and the smallest angles respectively. From the vertex that makes the largest angle, counting anti-clockwise the edges of the triangle until the vertex that makes the smallest angle, and these edge(s) ... 1 Because you can always make v = y \times a/|a| and then a \times v /|a| will give y, so x = v/|a| = y \times a/|a|^2 0 Write  y = y_1e_1+y_2e_2 + y_3e_3  and  a = a_1e_1 + a_2e_2 + a_3e_3 , where \{e_1,e_2,e_3\} is the canonical basis for \mathbb{R}^3. Consider the matrix$$ A = \begin{pmatrix} 0 & a_3 & -a_2\\ -a_3 & 0 & a_1\\ a_2 & -a_1 & 0 \end{pmatrix} $$and prove that the any non-trivial solution of Ax = y is the x you are looking ... 0 First of all (\nabla \times \vec{v})_i = \epsilon_{ijk}\partial_j v_k with Einstein summation. Then for any scalar \phi$$ (\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 $$the first one is because ... 1 n \perp P and v \parallel P, so n \perp v. Then L \in P and n \perp P, so n \perp L and v_1 \parallel L, so n \perp v_1. So \DeclareMathOperator{span}{span}n \in \span(v, v_1)^\top. Further v\times v_1 \perp v and v \times v_1 \perp v_1. This is a property of the vector product. So v \times v_1 \in \span(v, v_1)^\top. ... 0 @MvG has answered it and I accepted it. I just wanted to post a picture I've drawn for this to help me understand in case if someone else finds it useful. 1 Consider three-dimensional space \mathbb R^3. If you normally homogenize by appending z=1, that means that your geometry as you know it happens on the z=1 plane in space. But it's also possible to view geometric elements as linear (i.e. containing the origin) subspaces of the whole three-dimensional space. A point on the plane corresponds to a line ... 0 You can also use the Gram-Schimdt process. Knowing w and starting with a random vector u, you can find x by writing x = u - (w^T*u)/(w^T*w)*w. Matlab example: >> w = rand(4,1) w = 0.9575 0.9649 0.1576 0.9706 >> u = rand(4,1) u = 0.4218 0.9157 0.7922 0.9595 >> x = u - (w'*u)/(w'*w)*w x = -0.3755 ... 1 Assuming w_1 is non-zero: For any other w_j non-zero you will have a solution with x_1=1, x_j=-\dfrac{w_1}{w_j} and other x_k=0. You can then take combinations of such solutions to produce even more solutions. If any of the w_j are zero, you can also add to these solutions vectors where the corresponding x_js take any values. So in ... 1 Because you have a underdetermined equation (less equations than unknowns) you need to assign dummy variables. lets say our vecotr w=(w_1,w_2,w_3,w_4)^T. Then w^Tx=w_1+w_2x_2+w_3x_3+w_4x_4=0. We know that we have one equation and 3 unknows. Hence, we can choose 3-1=2 free variables. Let us call x_2=a and x_3=b. Now solve the equation for x_4. ... 1 Note that$$\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}.$$Since \vec{b} and \vec{c} are not parallel, \vec{a}\cdot\vec{b} = -\frac{\sqrt{3}}{2}, which implies that \|\vec{a}\|\|\vec{b}\|\cos\theta = -\frac{\sqrt{3}}{2}. Since \vec{a} and ... 0 Think thats IIT question hint a\times b\times c=(a.c)b-(a.b)c can you solve now? 2 Let i_x, i_y, i_z be the basis vectors of a right-handed coordinate system. Now we can write $$i_x \times i_y = i_z \quad i_y \times i_z = i_x \quad i_z \times i_x = i_y$$ By the definition of vector product if the vectors are orthogonal the sin (\theta) = 1 while if the vectors are parallel sin (\theta) = 0 then ... 2 First notice that if we define (u \wedge v ) \cdot w = \det (u,v,w) and let i,j,k,l = 1,2,3 then$$(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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