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| visits | member for | 2 years, 5 months |
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1d |
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Can you find the resultant force between these two vectors? @iostream007, I think your equation should be $|150+ 300\times \cos 110^\circ|$ because the negative will come from $\cos$. |
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1d |
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Circular motion trig No, because the argument to the $\sin$ and $\cos$ functions are $\theta = \dfrac{\pi}{3} - k \cdot 2\pi$ and $\theta = -\dfrac{\pi}{3} - k \cdot 2\pi$ which each only have one value in the interval $[0,2\pi]$. |
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May 9 |
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This is regarding Vector spaces So the space only has one element $a$? |
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Apr 26 |
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4 dimensional numbers Is your multiplication table right? By it, both the products $ij$ and $jk$ are anti-symmetric, however the product $ik$ is not. For example, $ik = i$, but $ki=-j$. Is this correct? |
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Apr 26 |
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Max of two vectors - how is this evaluated? It depends entirely on how you define $\max$. It is perfectly reasonable to define $\max \{\mathbb{v},0\}$ as being vector valued. In that case there is no ambiguity. For example, if $v = [-1, 1]^T$ then $\max \{\mathbb{v},0\} = [0, 1]^T$ |
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Apr 23 |
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What is this expression called? @Nexcius, You could think of $A_{1} * B_{2} - A_{2} * B_{1}$ as the determinant of a matrix or as a wedge product. |
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Apr 22 |
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Intuition why the volume and surface area of the unit sphere eventually decrease You should see the discussion here: math.stackexchange.com/questions/15656/… |
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Apr 22 |
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Determine all vector subspaces of the real vector space $\mathbb{R}^2$ The 2 dimensional subspace and 0 dimension subspace are trivial. How would you describe the set of 1 dimension subspaces? What would a basis of such a subspace look like? |
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Apr 22 |
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Is there a nice way to interpret this matrix equation that comes up in the context of least squares @crf, Once you reduce the problem to $A\mathbf{x}=\mathbf{y}$, you are breaking with the knowledge that the elements of $A$ come from powers of $x$. The expression of the normal equations basically projects the entire problem onto the column space of $A$. As long as the columns of $A$ are linearly independent, it does not matter where they came from. You are translating the problem from one domain (i.e. curve fitting) to another (linear algebra) to simplify the solution. You should not insist that notions from one domain maintain their meaning in the other. |
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Apr 19 |
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how to i answer this calculus hw problem? How would you approach the problem with only one sub-interval (i.e. [1,5])? |
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Apr 3 |
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Cross Product Intuition +1: This is a great question! |
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Mar 23 |
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An eigenvector is a non-zero vector such that… Yes, but they are defined that way precisely because they represent a non-trivial subspace (i.e. they have to span that space). In other words... the feature that makes them interesting and worthy of definition is that they are non-zero and still satisfy $\mathbf{Ax}=\lambda\mathbf{x}$. |
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Mar 23 |
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An eigenvector is a non-zero vector such that… @ScottH. I think that you are confusing the notions of a space and a basis. A basis is a non-zero vector which is linearly independent (i.e. cannot contain the zero vector) and spans the space. An eigenvector cannot be zero because it is a basis. The space spanned by the eigenvector must contain the zero vector. |
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Mar 22 |
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An eigenvector is a non-zero vector such that… An eigenspace is spanned by a non-zero eigenvector associated with a particular eigenvalue. The eigenspace must be at least one dimensional and therefore excludes using the zero vector as an eigenvector. |
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Mar 6 |
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New vector position Is the axis aligned with one of your coordinate axes. If so, then you can omit that axis in your calculations. Don't include it in your distance calculations and don't include it in your scaling calculations. If it is not aligned with a coordinate axis, then you need something more sophisticated. |
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Mar 6 |
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New vector position It is essentially the same equation just written a different way. The steps you need to go through are as follows: (1) compute the average distance, (2) compute the relative vector to the center $v_i - c$, (3) normalize it to unit length by dividing it by its own length, (4) scaling it to the average length and then finally (5) adding the center back onto the result. The referenced post does essentially the same thing, except that it distributes the scaling to each component. |
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Mar 5 |
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New vector position Almost. You should use: npX = (vts[0] - cs[0]) / (distanceFromCenter[for this point]) * averageDistance + cs[0] |
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Mar 5 |
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New vector position Just the magnitude of the vector. I believe that you compute it in your first loop and call it distanceFromCenter. |
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Mar 5 |
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New vector position So you have 8 points (for example) and a center point. You want to update the 8 points based on their distance to the center point. It is not clear what you want the new points to be based on their distance to the center point. Please clarify |
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Mar 5 |
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New vector position observing your code, it appears that you are summing the coordinates in oldCoordArray. If you divide by the number of points, you should have the center of the points. |
