As the title says, how would I obtain the vector $\mathbf t$ if the vector $\mathbf i$ and the refractive indices are known? I've tried for hours but I keep getting stuck. Is there an equation readily available? Thank you.
122663
Snell's law of refraction at the interface between 2 isotropic media is given by the equation: \begin{equation} n_1 \,\text{sin} \,\theta_1 = n_2 \, \text{sin}\,\theta_2 \end{equation} where $\theta_1$ is the angle of incidence and $\theta_2$  the angle of refraction. $n_1$ is  the refractive index of the optical medium in front of the interface and $n_2$ is the refractive index of the optical medium behind the interface.  How can this be expressed in vector form: \begin{equation} n_1(\textbf{i} \times \textbf{n}) = n_2 (\textbf{t} \times \textbf{n})  \end{equation} where $\textbf{i}(i_\text{x}, i_\text{y}$ and $i_\text{z})$ and  $\textbf{t}(t_\text{x}, t_\text{y}, t_\text{z})$  are  the unit directional vector of the incident and transmitted ray respectively.   $\textbf{n}(n_\text{x}, n_\text{y}, n_\text{n})$ is the unit normal vector  to the interface  between the two media pointing from medium 1 with refractive index $n_1$ into medium 2 with refractive index $n_2$.    Further more how can the Snell's law of refraction be expressed$^\text{1}$ in the following way?  \begin{equation} t = \mu \textbf{i} + n\sqrt{1- \mu^2[1-(\textbf{ni})^2]} - \mu \textbf{n}(\textbf{ni}) \end{equation} Here $\mu = \dfrac{n_1}{n_2}$ and $\textbf{n}\textbf{i}= n_{\text{x}} i_{\text{x}} + n_{\text{y}}i_{\text{y}}  + n_{\text{z}} i_{\text{z}}$ denotes the dot (scalar) product of vectors $\textbf{n}$ and $\textbf{i}$.  References:   Antonín Mikš and Pavel Novák, Determination of unit normal vectors of aspherical surfaces given unit directional vectors of incoming and outgoing rays: comment, 2012 Optical Society of America, page 1356
61910
I understand the equation explaining angles of refraction related to wavelengths. But I'm looking for a more intuitive approach/visual explanation? Perhaps related to QM?
30268
$$\mathbf D = \varepsilon \mathbf E$$ I don't understand the difference between $\mathbf D$ and $\mathbf E$.  When I have a plate capacitor, a different medium inside will change $\mathbf D$, right? $\mathbf E$ is only dependent from the charges right?
85459
In classic OLS regression it is well-known that $(\mathbf{I}-\mathbf{H})\mathbf{y}=\mathbf{r}$, where $\mathbf{I}$ is the identity matrix, $\mathbf{H}$ is the , $\mathbf{y}$ is the vector of response (dependent) values, and $\mathbf{r}$ is the vector of residuals.  My question is whether the scalar $\mathbf{y}^T(\mathbf{I}-\mathbf{H})\mathbf{y}=\mathbf{y}^T\mathbf{r}=\mathbf{y}^T\mathbf{y}-\mathbf{y}^T\mathbf{\hat{y}}$ has some meaningful interpretation.
36329
Question: Let $v \in \mathbb{R}^3$ be a unit vector. If $\{v_1, v_2, v_3\}$ is an orthonormal basis of $\mathbb{R}^3$ and $v = a_1v_1 + a_2v_2 + a_3v_3$, can we have $|a_1| &gt; 3$?  What I think:  Since the norm of $v$ is one, then   $\sqrt{v\cdot v} = 1$  $(a_1v_1)^2 + (a_2v_2)^2 + (a_3v_3)^2 = 1^2$  Since $3^2 = 9 $ and $v_1 \cdot v_1 = 1^2$ (since each vector's norm in an orthonormal basis is 1), then $(a_1v_1) \geq 9$. Since $(a_2v_2)^2 + (a_3v_3)^2 &gt; 0,$ then $(a_1v_1)^2 + (a_2v_2)^2 + (a_3v_3)^2 &gt; 1$. Hence, $|a_1| &lt; 3.$  I'm pretty sure my assumption of $(a_1v_1) \geq 9$ is wrong but I don't know how else to solve this problem.
65812
I'm trying to compute the shadow of 3D objects on the ground. To do this I'm assuming parallel rays from Sun origin.   Let's assume the Sun direction is given by azimuth $\varphi$ and elevation $\theta$ angles. So the direction unit vector is retrieved as:  $$ \hat{\mathbf{s}} = \mathbf{R}_y\left(\theta\right)\mathbf{R}_z\left(\varphi\right)\hat{\boldsymbol{\imath}}$$  where $\mathbf{R}_{x/y/z}$ are the rotation matrices along the $x, y, z$ axes given by, for example, Euler-Rodrigues equation, and $\hat{\boldsymbol{\imath}}$ is the unit vector along $x$-axis.  A point on a ray-line has the equation:  \begin{equation} \mathbf{r} = \mathbf{r}_p + \lambda\hat{\mathbf{s}}  \tag{1} \end{equation}  being $\mathbf{r}_p$ a point on that line.  A point on a general plane is given by the equation:  \begin{equation} \left(\mathbf{r} - \mathbf{r_0}\right)\cdot \hat{\mathbf{n}} = 0 \tag{2} \end{equation}  being $\mathbf{r}_0$ a point on the plane and $\hat{\mathbf{n}}$ the unit vector normal to that plane.   Ensuring the validity of both equations (1) and (2) allows to compute the $\lambda$ multiplier and retrieve the position vector of the projected point (writing $\tilde{\mathbf{r}} = \mathbf{r_p} - \mathbf{r_o}$)  \begin{gather} \lambda = - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right) \\ \mathbf{r}_\text{proj} = \mathbf{r}_p - \left(\frac{\tilde{\mathbf{r}}\cdot \hat{\mathbf{n}}} {\hat{\mathbf{s}}\cdot\hat{\mathbf{n}}}\right)\hat{\mathbf{s}} \tag{3} \end{gather}  Applying eq. (3) to all the vertices of a mesh $\left\{\mathbf{r}_{p_i}\right\}$ .   A problem can be noted: $\hat{\mathbf{s}}\cdot \hat{\mathbf{n}}$ can be zero (e.g. Sun is horizontal, i.e. the ray is ortogonal to the normal of plane).   Are there some conceptual mistakes in my reasoning? Is there a way to rewrite the operation as a linear operator applied to the set of vertices position vectors?   $$ \left[\mathbf{A}\right]\left\{\mathbf{r}_{p_i}\right\} = \mathbf{r}_\text{proj}$$
48051
It's not immediately clear to me why this is true. My notes say that putting $n$ orthonormal vectors $ v_1, ..., v_n$ in the columns of $X$ gives $X^TX = I$, and it follows from this that the rows of $X$ are orthonormal.   Can you explain this better to me? I tried the $2\times 2$ case, but it isn't clear to me why this is true.
71848
I was trying to derive Lagrangian for the electromagnetic field from Lorentz's force formula $$\mathbf{F}= q(\mathbf{E} + \mathbf{v}\times \mathbf{B})$$ I have to find the potential by using the line integral of the force. I have written the above equation in terms of scalar and vector potential which is given by. $$F=q(-\nabla\phi-\frac{1}{c}\frac{d\mathbf{A}}{dt}+\frac{1}{c}\nabla(\mathbf{v}\cdot \mathbf{A}))$$ now I have to find potential by using line integral $$V=-\int \mathbf{F}\cdot d\mathbf{r}$$ now I have to show that the integral of the second term is zero so that I can get exact potential which is given by$$V=q(\phi-\frac{1}{c}\mathbf{v}\cdot \mathbf{A})$$ So my question is how to show that the integration of second term is zero?
17782
Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$.  a) Find the base and dimension of fundamental subspaces of matrix A  b) Using the results given by a) find the general solution for $Ax - z_1$.  Any hint would be appreciated.
50539
What is the physical significance of  $$\nabla\times\boldsymbol{V}~?$$  I mean I read 'curl V represents the rotation of the vector $V$. My question what is it about the term $\nabla\times\boldsymbol{V}$ that it represents the rotation of the vector?
63164
I don't think I ever understood the rationale behind this.  I get that the dot product $\mathbf{a} \cdot  \mathbf{b} =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b}\rVert \cos\theta$ is derived from the cosine rule. (Do correct me if I'm wrong.)  However, I never really understood why $\mathbf{a} \times  \mathbf{b} =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b} \rVert \sin\theta$. Is there a proof for this?  In addition, how does the vector cross product lead to orthogonal vectors?  I remember learning that they do and I know how to solve equations using the respective formulas, but I never got why is it so.
110433
Let $\vec{k}$, $\vec{v}$, and $\vec{u}$ be vectors, where $\vec{u}$ is unknown and $\vec{k}$ and $\vec{v}$ are known vectors. Given:     $\vec{u}\cdot\vec{k}=c$  $\vec{u} \times \vec{k}= \vec{v}$   From this relations, how can i determine the vector $u$? I tried to construct orthogonal coordinate system from $(\vec{k}, \vec{v},\vec{k}\times \vec{v})$ but i couldn't proceed from there. Any idea?
83790
How do you find a vector in the form  when only the angle and magnitude are given?  Here is an example where an angle of 80 degrees is given along with a magnitude of 3.
79351
I learned vector analysis and multivariate calculus about two years ago and right now I need to brush it up once again. So while trying to wrap my head around different terms and concepts in vector analysis, I came to the concepts of vector differentiation, gradient, divergence, curl, Laplacian etc.   The reference that I'm using is very inadequate to give any geometric/physical interpretetions of these (almost) new concepts. I don't have much problem with their formulae and working rules, but I want to look at them in a little more geometric way. For example, the geometric significance of gradient that I got out of my book is the following: if $f:\mathbb{R}^3\to \mathbb{R}$ be a differentiable function then $\nabla f (x,y,z)$ is the vector perpendicular to the level surface $f(x,y,z)=c$ ($c$ constant) at the point $(x,y,z)$.  I would really appreciate if anyone can explain how, in this way, can divergence, curl and Laplacian be interpreted geometrically. (e.g. For a given vector field $\textbf{F}: \mathbb{R}^3\to \mathbb{R}^3$, the relation between $\nabla\times\textbf{F}$ and $F$ and questions like that.) I looked around Google a bit, but couldn't find what I was looking for.  Thanks in advance.
86114
So I have this problem (self learning linear algebra) and I try to solve it using the indicated steps. Question is: Am I right, is this a valid answer? Obviously for me makes sense but since I'm just starting with maths maybe there's something wrong:     If $\|\vec{v}\| = 5$ and $\|\vec{w}\| =3$, what are the smallest and largest values of $\|\vec{v}-\vec{w}\|$ and $\vec{v} \cdotp \vec{w}$?   EDIT   I'd moved my initial answer to a full answer. See below.
18435
Given a column vector $v$ in $\mathbb{R}^n$, what are the eigenvalues of matrix $vv^T$ and associated eigenvectors?  PS: not homework even though it may look like so.
121069
I feel like this should be straightforward, but does anyone have a proof of the following?  Let $f: \mathbb{R}^n \to \mathbb{R}$ satisfy the following. For each coordinate $i$, for an arbitrary vector $x_{i}$, define $ f_i(y)$ to be $f$ restricted to the $i^{th}$ parameter, fixing the others $ x_{i}$; then $ f_i(y)$ is convex. Then $f$ is a convex function.  Also, if there is a more standard math notation I should use to describe this problem, what would that be?  (Edit: or a counterexample...)  Thanks!    I should add what I've tried.  $g(\lambda \vec{x} + (1-\lambda)\vec{y}) = g(\lambda x_1 + (1-\lambda)y_1,\dots,\lambda x_n + (1-\lambda)y_n)$  $\leq \lambda g(x_1,\lambda x_2 + (1-\lambda)y_2,\dots) + (1-\lambda) g(y_1,\lambda x_2 + (1-\lambda)y_2,\dots)$  I don't see this going anywhere good, we end up with something nasty like  $\lambda^n g(\vec{x}) + \lambda^{n-1}(1-\lambda) \sum_i g(y_i,\vec{x}_{i}) + \dots + (1-\lambda)^n g(\vec{y})$  I think. Maybe we can get it from there?