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There is a sense where you can start with just a scalar product. If you assume that $v^2=vv=v\cdot v=|v|^2$ holds for any vector then that is a scalar product for a vector with itself. From that you can get the scalar product of any two vectors $v$ and $w$, $v \cdot w= \frac{1}{2}\left((v+w)^2-v^2-w^2\right)=\frac{1}{2}\left(vw+wv\right)$ where for the ...


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In short: The dot product gives you the multiplication of the parallel components. Example: The work expression $W=\vec F \cdot \vec r$, where only the force component parallel to the direction (or likewise, the position component parallel to the force) is wanted. The (magnetude of the) cross product gives you the multiplication of the perpendicular ...


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The dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. This operation can be defined either algebraically or geometrically. The cross product or vector product is a binary operation on two vectors in three-dimensional space and is denoted by the symbol ×. The cross product a ...


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The right hand rule and cross product of matrices is a convenient and useful tool for describing natural phenomena in three dimensional space. Like all conceptual tools, it was devised by humans to acquire knowledge, and to further our ability to predict and to manipulate aspects of our world. We use it because it fits with nature. Examples ($\times$ ...


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Why should the resultant cross-product necessarily point in the orthogonal direction given by the right hand rule? Because, we made it to. Other math tools give other results. We made them to do that too. We could make any mathematical method to bring any result we want. But some methods, tools and definitions turn out to fit with the real world. ...


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Here we are talking about instantaneous velocity. So,its -20 m/s. And the velocity will be always tangent to the circular track. So,it will be 20 m/s in magnitude every second. But the direction will be different so different values in x & y - axes


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I'm not knowledgeable in some aspects of the question, but I will provide an answer unrelated to others. An object can rotate so fast that some representations of angular velocity cannot be valid. For example, when an object rotates more than 180-degrees or 360-degrees (pi or 2*pi radians) per unit time, the representation must be able to represent such ...


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Displacement refers to the object's position relative to the observer. The "place in space" of the orange. Distance is the object's position relative to an earlier position. If you pick up an orange, and run 10 miles holding it straight out, no work gets done on the orange. But if you extended your arm 10 miles, you would have to be doing work on the ...


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It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...


3

If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ...


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Everything depends on how your fields (vectors and spinors are fields in the classical theory, and when you quantize in QFT, they become operator-valued fields) transform when you make a Lorentz transform: An scalar is a field that doesn't change at all: $\phi'(x') = \phi(x)$. Examples are the Higgs and pions. A vector field is a field that transform like ...


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No, you should not write "$\left|\vec{F}\right| = 30\textrm{N}$", because it's no better than "$F=30\textrm{N}$" Since force is a vector, you could write out the list of components, either as a parenthetical list or a column vector: $$\vec{F} = \left(30\textrm{N}\right) = \left[ 30\textrm{N}\right]$$ You could also write the one component as a scalar: ...


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Some people use $\mathbf{F}$ instead of $\vec{F}$ or even $\overrightarrow{F}$. I agree that often $F=\| \vec{F} \|$ is a convenient shortcut. So for example A force $\mathbf{F}=(10 \mbox{ N},0,0)$ has magnitude $\|\mathbf{F}\|=10 \mbox{ N}$. The components of $\mathbf{F}$ are $F_x = 10\mbox{ N}$, $F_y=0$ and $F_z=0$ So the subscript is used to ...


6

Force is indeed a vector. Technically you should write $|\overrightarrow{F}| = 30N$, however there is usually context given that let you omit this. If you are working in one dimension, then the vector-like direction is all encapsulated in the sign once you've defined your coordinate system (e.g. -30N is 30N downwards.) Beyond that, it is typically just a ...


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A negatively charged particle has an electric field, $$\mathbf{E} =-\vert\mathbf{E}\vert \, \hat{\mathbf{r}} =\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}$$ Gauss's law gives, \begin{align*} \int \mathbf{E}\cdot d\mathbf{A} &= \int \Big(\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}\Big) \cdot (r^2 \sin \theta d\theta d\phi \, \hat{\mathbf{r}}) = ...


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In mechanics the vector cross product is used to transform a force at a distance into a moment, and a rotation about an axis into linear velocity $$ \mathbf{M} = \mathbf{r} \times \mathbf{F} \\ \mathbf{v} = \mathbf{r} \times \mathbf{\omega} $$ If you project those vectors into any plane, you will see that each component of force (or rotation) is multiplied ...


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A vector product of two vectors is also called a cross product. As you may know, one way of describing a vector, as opposed to defining a vector is "a quantity which has magnitude and direction." In 3-dimensional space, one can specify a vector as the sum of $x, y,$ and $z$ components of parts. We can write $$\vec{A} = (A_x, A_y, A_z) \text{ or } ...


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A$\times$B is a vector that is perpendicular to both A and B, and its norm (length) is equal to the area of the parallelogram whose sides are the vectors A and B. You can check that that seemingly random expression satisfies both properties.


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When the slope is constant (left part of the drawing with the skier, flatness in the sled example) any possible acceleration is tangent to the surface. Thus, when the sled is speeding along perfectly level and flat ground, it experiences no acceleration. When the skier is going downhill, he experiences only acceleration along the slope. But, as already ...


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In modern mathematical terminology, a functor is called covariant when it preserves the direction of the morphisms, contravariant if it reverses it. For a given differentiable map between manifolds (of which a special case would be open sets within the same manifold), the derivative is a map between the associated tangent bundles. This defines a covariant ...


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Since momentum is a vector, the quantity being measured did indeed change. As you state, while the magnitude is a constant $10\ \mathtt{kg\cdot m/s}$, the direction has altered. But what does that mean? The meaning of this change lies in understanding the distinction between momentum and impulse. The change of momentum is called an impulse. The common ...


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$$\begin{align} \left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}\epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = ...


6

I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$ ...


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It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.


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A vector field is a function which associates to a point of the space (domain) a vector, namely a set of number. So, in this case, a well defined magnetic field vector gives you these sets of numbers for every point of the space. Each of these sets of numbers contains information about the intensity and the direction of the magnetic field in the relative ...


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This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.


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A vector is a quantity having a direction and a magnitude. Magnets have a magnetic influence, which can be visualized a physical vector field. A magnetic field vector allows you to predict the influence of a magnet on a magnetic material.



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