Vector Examples, Equations, and Practice Problems

I have seen that sometimes the Navier-Stokes equations are written with the term $(\mathbf{v}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->})\mathbf{v}$ expressed as $\mathbf{v}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{v}$. However, is it true in general the following equality for any vector field $\mathbf{v}$?
$(\mathbf{v}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->})\mathbf{v}\stackrel{?}{=}\mathbf{v}\cdot <!--\; \cdot \; -->(\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{v})$
A couple of examples: in Batchelor's An Introduction to Fluid Dynamics, equation (2.1.2) defines the mass derivative of the velocity field (which is the LHS of the NS equation) as
$\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{u}}{\mathrm{\partial <!--\; \partial \; -->}t}+\mathbf{u}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{u}$
Whereas, in Landau-Lifschitz Fluid Mechanics, 2nd edition, the Navier-Stokes equation is written in equation (15.7) as
$\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{v}}{\mathrm{\partial <!--\; \partial \; -->}t}+(\mathbf{v}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->})\mathbf{v}=-<!--\; -\; -->\frac{1}{\mathrm{\rho <!--\; \rho \; -->}}\mathrm{\nabla <!--\; \nabla \; -->}p+\frac{\mathrm{\eta <!--\; \eta \; -->}}{\mathrm{\rho <!--\; \rho \; -->}}\mathrm{\Delta <!--\; \Delta \; -->}\mathbf{v}$

Find the intersection of a line formed by the intersection of two planes $\stackrel{\to <!--\; \to \; -->}{r}.{\stackrel{\to <!--\; \to \; -->}{n}}_1=p_1$ and $\stackrel{\to <!--\; \to \; -->}{r}.{\stackrel{\to <!--\; \to \; -->}{n}}_2=p_2$
I know that the line would be along $({\stackrel{\to <!--\; \to \; -->}{n}}_1\times <!--\; \times \; -->{\stackrel{\to <!--\; \to \; -->}{n}}_2)$. So i need a point on the line to get the equation. I assumed a point C such that $\stackrel{\to <!--\; \to \; -->}{OC}$ is perpendicular to the line of intersection. I dont really know how to proceed from here. Do I have to use the equations $\stackrel{\to <!--\; \to \; -->}{c}.{\stackrel{\to <!--\; \to \; -->}{n}}_1=p_1$ and $\stackrel{\to <!--\; \to \; -->}{c}.{\stackrel{\to <!--\; \to \; -->}{n}}_2=p_2$?

Let $\mathrm{\lambda <!--\; \lambda \; -->}$ be an eigenvalue of A, such that no eigenvector of A associated with $\mathrm{\lambda <!--\; \lambda \; -->}$ has a zero entry. Then prove that every list of n−1 columns of $A-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}I$ is linearly independent.

Name of a vector of 1s?
Let's consider an N dimensional vector where each coordinate takes the value 1. For example, for $N=5$ we have: $(1,1,1,1,1)$
Does this type of vector have a name in the literature? Perhaps "unary?". Also, are there any conventions on how to refer to it in terms of notation? (e.g. ${\mathbf{1}}^T$?)

If I use the rule of vector triple product, it becomes:
$\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{a}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b}=\stackrel{\to <!--\; \to \; -->}{a}(|\stackrel{\to <!--\; \to \; -->}{b}{|}^2)-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}(\stackrel{\to <!--\; \to \; -->}{b}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{a})$
which is generally non-zero, but suppose I use properties of cross product:
$\stackrel{\to <!--\; \to \; -->}{a}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b}=-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{a}$
Hence,
$\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{a}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b}=-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b}=\stackrel{\to <!--\; \to \; -->}{a}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{b})=0$
What did I do wrong?

Can we solve $\mathit{a}\mathit{P}=\mathit{b}$ where $\mathit{a}$ and $\mathit{b}$ are each $1\times <!--\; \times \; -->n$ row vectors and are known, and $\mathit{P}$ is an $n\times <!--\; \times \; -->n$ permutation matrix that is unknown?

We have the vectors $v=i+j+2k=(1,1,2)$ and $u=-<!--\; -\; -->i-<!--\; -\; -->k=(-<!--\; -\; -->1,0,-<!--\; -\; -->1)$
I want to calculate the angle between u and v in radians using the cross product.
I have done the following:
$\begin{array}{rl}|v\times <!--\; \times \; -->u|=|v|\cdot <!--\; \cdot \; -->|u|\cdot <!--\; \cdot \; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}& \Rightarrow <!--\; \Rightarrow \; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{|v\times <!--\; \times \; -->u|}{|v|\cdot <!--\; \cdot \; -->|u|}=\frac{\sqrt{3}}{\sqrt{6}\cdot <!--\; \cdot \; -->\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{3}\cdot <!--\; \cdot \; -->\sqrt{2}\cdot <!--\; \cdot \; -->\sqrt{2}}=\frac{1}{2}\\ & \Rightarrow <!--\; \Rightarrow \; -->\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}n+\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}\text{ or }\mathrm{\theta <!--\; \theta \; -->}=2\mathrm{\pi <!--\; \pi \; -->}n+\frac{5\mathrm{\pi <!--\; \pi \; -->}}{6},n\in <!--\; \in \; -->\mathbb{Z}\end{array}$
Is everything correct?
Which of the values do we choose? Or are both valid?

Consider for example the notation $\mathbf{x}\in <!--\; \in \; -->{\mathbb{Z}}^4$. This could mean that x contains duplicate values, e.g. $\mathbf{x}=\{0,0,1,2\}$. Is there any way I can express that x is a vector of integers without any duplicates, e.g. $\mathbf{x}=\{5,4,3,2\}$? Or should I just mention it in the text?

How does one write the following expression
$D_{jk}(r_k{\mathrm{\delta <!--\; \delta \; -->}}_{ij}-<!--\; -\; -->r_i{\mathrm{\delta <!--\; \delta \; -->}}_{jk}-<!--\; -\; -->r_j{\mathrm{\delta <!--\; \delta \; -->}}_{ik})$
in matrix notation? Is this just
$\mathbf{\text{D}}(\mathbf{\text{r}}\times <!--\; \times \; -->\mathbf{\text{I}})$?

The conditions $\stackrel{\to <!--\; \to \; -->}{r}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{s}=0$, and $\stackrel{\to <!--\; \to \; -->}{r}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{x}=c$, and $\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{x}=\stackrel{\to <!--\; \to \; -->}{s}$ . Find x in each of the three mutually orthogonal directions, $\stackrel{\to <!--\; \to \; -->}{r}$ , $\stackrel{\to <!--\; \to \; -->}{s}$ , and $\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}$
So far $\stackrel{\to <!--\; \to \; -->}{x_r}=\frac{\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{r}}{|r|}\frac{\stackrel{\to <!--\; \to \; -->}{r}}{|r|}=\frac{c}{|r{|}^2}\stackrel{\to <!--\; \to \; -->}{r}$ , and $\stackrel{\to <!--\; \to \; -->}{x_s}=\frac{\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{s}}{|s|}\frac{\stackrel{\to <!--\; \to \; -->}{s}}{|s|}=\frac{\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{x})}{|s|}\frac{\stackrel{\to <!--\; \to \; -->}{s}}{|s|}=0$ Since $\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{x})$ is the volumn of a degenerate parallelepiped. Where I'm having most difficulty is in...
$\stackrel{\to <!--\; \to \; -->}{x_{\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}}}=\frac{\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s})}{|\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}{|}^2}\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}$ . Since r and s are orthogonal does that mean $|\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}|=|r||s|$. And also can I calculate $\stackrel{\to <!--\; \to \; -->}{x}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s})$using the triple product to be $|r||s||\stackrel{\to <!--\; \to \; -->}{x_{s\perp <!--\; \perp \; -->}}|=|r||s||\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{x_s}|=|r||s||\stackrel{\to <!--\; \to \; -->}{x}|$. Is there a simpler simplification of this?
So does $\stackrel{\to <!--\; \to \; -->}{x_{\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{s}}}=\frac{|\stackrel{\to <!--\; \to \; -->}{x}|}{|\stackrel{\to <!--\; \to \; -->}{r}||\stackrel{\to <!--\; \to \; -->}{s}|}\stackrel{\to <!--\; \to \; -->}{x}$?

Let $F(x,y)=⟨<!--\; ⟨\; -->-<!--\; -\; -->y,x⟩<!--\; ⟩\; -->$ and C be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ oriented counter clockwise, then find the value of ${\int <!--\; \int \; -->}_{C}F.dr$
This is how I tried it,
$x=4\cos <!--\; \; -->(t)\to <!--\; \to \; -->dx=-<!--\; -\; -->4\sin <!--\; \; -->(t)$
$y=3\sin <!--\; \; -->(t)\to <!--\; \to \; -->dy=3\cos <!--\; \; -->(t)$
and $0\le <!--\; \le \; -->t\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$
${\int <!--\; \int \; -->}_{C}F.dr=\int <!--\; \int \; -->-<!--\; -\; -->ydx+xdy={\int <!--\; \int \; -->}_{t=0}^{2\mathrm{\pi <!--\; \pi \; -->}}12({\sin }^2<!--\; \; -->(t)+{\cos }^2<!--\; \; -->(t))dt=75.4$
But answer given to me at the back is 98.2 and doesn't agrees with mine.

I was given the following definition and proposition.
Definition: The set $\{x\in {\mathbb{R}}^n|Ax=0\}$ is the null set of A.
Proposition: If p is a vector such that Ap=b, then $\{x\in {\mathbb{R}}^n|Ax=b\}$ = $\{y+p|y\in NS(A)\}$
I am just so perplexed by the sudden introduction of notations and all of the null set.
Can someone please explain this in simple terms and explain what the notations are?

Consider the plane, 𝒫 in ${ℝ}^3$ by the vector equation
$x(s,t)=(1,-<!--\; -\; -->1,2)+s(1,0,1)+t(1,-<!--\; -\; -->1,0);s,t\in ℝ$
Compute a unit normal vector, n, to this plane.
My attempt is the third normal vector is $(1,\frac{2s}{t}+1,1)$ and the unit normal vector I got is
$\frac{1}{\sqrt{3+\frac{4s^2}{t^2}+\frac{4s}{t}}}(1,\frac{2s}{t}+1,1)$

I was working on two Examples of Friedberg- Insel-Spence's Linear Algebra. In example 6, in ${\mathbb{R}}^2$(not 2-dimensional real vector space, consider it as the set $\mathbb{R}\times <!--\; \times \; -->\mathbb{R})$ scalar multiplication was defined as usual, but vector addition was defined as the following:
$(a_1,b_1)+(a_2,b_2)=(a_1+b_2,a_1-<!--\; -\; -->b_2)$ for any $(a_1,b_1),(a_2,b_2)\in <!--\; \in \; -->{\mathbb{R}}^2.$ The set ${\mathbb{R}}^2$ is closed addition and scalar multiplication, but it's not a vector space over $\mathbb{R}$ because ${\mathbb{R}}^2$ is not an abelian group under addition, for instance, this operation is neither commutative nor associative. Moreover, there is an issue with the distribution. Is there any complete list of ways(addition + scalar multiplication) such that the set ${\mathbb{R}}^2$ is a 2-dimensional vector space?

I'm given that
$\stackrel{\to <!--\; \to \; -->}{E}=\frac{{\mathrm{\mu <!--\; \mu \; -->}}_0{p}_0{\mathrm{\omega <!--\; \omega \; -->}}^2}{4\mathrm{\pi <!--\; \pi \; -->}r}[\cos <!--\; \; -->u(\stackrel{^<!--\; ^\; -->}{x}-<!--\; -\; -->\frac{x}{r}\stackrel{^<!--\; ^\; -->}{r})+\sin <!--\; \; -->u(\stackrel{^<!--\; ^\; -->}{y}-<!--\; -\; -->\frac{y}{r}\stackrel{^<!--\; ^\; -->}{r})]$
implies
$-<!--\; -\; -->\frac{{\mathrm{\mu <!--\; \mu \; -->}}_0{p}_0{\mathrm{\omega <!--\; \omega \; -->}}^2}{4\mathrm{\pi <!--\; \pi \; -->}r}[\cos <!--\; \; -->u\stackrel{^<!--\; ^\; -->}{x}\times <!--\; \times \; -->\stackrel{^<!--\; ^\; -->}{r}+\sin <!--\; \; -->u\stackrel{^<!--\; ^\; -->}{y}\times <!--\; \times \; -->\stackrel{^<!--\; ^\; -->}{r}]=\frac{1}{c}\stackrel{^<!--\; ^\; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{E},$
but I don't follow how to get from the former to the latter.

I have two high dimensional unit vectors $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$ and the distance d between them.
I want to change the vector $\stackrel{\to <!--\; \to \; -->}{a}$ so that it is still a unit vector but the distance to $\stackrel{\to <!--\; \to \; -->}{b}$ is now $d^{\prime }$.
I already tried it this way:
${\stackrel{\to <!--\; \to \; -->}{a}}^{\prime }=\stackrel{\to <!--\; \to \; -->}{a}+(\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}\ast <!--\; \ast \; -->(1-<!--\; -\; -->\frac{d^{\prime }}{d}))$
But then the new vector does not have unit length.
Anyone any idea?

How to solve the equation and directly determine the value of $\stackrel{\to <!--\; \to \; -->}{v}$ in the equation?
$\stackrel{\to <!--\; \to \; -->}{v}-<!--\; -\; -->a[\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{^<!--\; ^\; -->}{y}]=b\stackrel{\to <!--\; \to \; -->}{E}$
where, $\stackrel{\to <!--\; \to \; -->}{v}$ and $\stackrel{\to <!--\; \to \; -->}{E}$ are in the $\stackrel{^<!--\; ^\; -->}{x}$ direction, and a and b are scalars.

Given the Function:
$f:{\mathbb{R}}^2\to <!--\; \to \; -->{\mathbb{R}}^2$ , $r\in <!--\; \in \; -->\mathbb{R}.$
$f(x,y)=\{\begin{array}{l}x=r\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\\ y=r\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})\end{array}$
Calculate this Partial Derivative:
$\frac{\mathrm{\partial <!--\; \partial \; -->}(x,y)}{\mathrm{\partial <!--\; \partial \; -->}(r,\mathrm{\theta <!--\; \theta \; -->})}$
I do really need some help on this lads, any help would be really appreciated.

3 points ABC on a plane, O as origins. $OA=\stackrel{\to <!--\; \to \; -->}{a}$ , $OB=\stackrel{\to <!--\; \to \; -->}{b}$, $OC=\stackrel{\to <!--\; \to \; -->}{c}$ . Point M inside $\mathrm{△<!--\; △\; -->}ABC$. And $\mathrm{△<!--\; △\; -->}MAB:\mathrm{△<!--\; △\; -->}MBC:\mathrm{△<!--\; △\; -->}MCA=2:3:5$
A straight line BM intersect side AC at N. Express OM in terms of vector a,b,c.
Can you give me some hint? I have been thinking, what i got is, $BM:MN=1:1.$ $AB:BC=2:3$ $\stackrel{\to <!--\; \to \; -->}{A}B=OB-<!--\; -\; -->OA=\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}$ $\stackrel{\to <!--\; \to \; -->}{C}B=OB-<!--\; -\; -->OC=\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}$
$\stackrel{\to <!--\; \to \; -->}{O}M=OB+BM$
$\stackrel{\to <!--\; \to \; -->}{B}M=\frac{1}{2}BN$ Then, i have difficulity in expressing $BN$ in terms of vector a,b,c.

What does mean $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{e_1,-<!--\; -\; -->e_l,\dots <!--\; \dots \; -->,e_d,-<!--\; -\; -->e_d\}$ and $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{\{+1,-<!--\; -\; -->1{\}}^d\}.$
I could not understand, need a simple explanation.