Vector Examples, Equations, and Practice Problems

Under which of the following conditions will the points A,B,C with position vectors $\stackrel{\to <!--\; \to \; -->}{a}$ , $\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{c}$ respectively be collinear?
(a) $\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}=2(\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a})$
(b) $|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}|=2|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}|$
(c) $\stackrel{\to <!--\; \to \; -->}{a}=2(\stackrel{\to <!--\; \to \; -->}{b}+\stackrel{\to <!--\; \to \; -->}{c})$
(d) $2\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}=\stackrel{\to <!--\; \to \; -->}{c}$
(e) $3\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->2(\stackrel{\to <!--\; \to \; -->}{b}+\stackrel{\to <!--\; \to \; -->}{c})=\stackrel{\to <!--\; \to \; -->}{0}$
My Try
Since (a) yields $\stackrel{\to <!--\; \to \; -->}{AC}=\mathrm{\lambda <!--\; \lambda \; -->}\stackrel{\to <!--\; \to \; -->}{AB}$, it is definitely collinear.
Furthermore I know that magnitude does not affect collinearity. I'm stuck determining (c),(d),(e).

Given
$u_m(x)=\frac{8}{R^3}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}{m}_1{x}_1}{R}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}{m}_2{x}_2}{R}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}{m}_3{x}_3}{R}$
with the boundary $\mathrm{\delta <!--\; \delta \; -->}\mathrm{\Omega <!--\; \Omega \; -->}=\{x\mid <!--\; \mid \; -->x_1=R,0<x_2<R,0<x_3<R\}$ and m is a vector with $m=(m_1,m_2,m_3),m_i\in \mathbb{N}$
Then calculates the normale derivative $\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}n}{u}_m(x)$ on the boundary $\mathrm{\delta <!--\; \delta \; -->}\mathrm{\Omega <!--\; \Omega \; -->}$ and derives
$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}n}{u}_m(x)=\frac{8\mathrm{\pi <!--\; \pi \; -->}{m}_1}{R^4}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}{m}_2{x}_2}{R}\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}{m}_3{x}_3}{R}\end{array}$
I'm very new to the topic of normal derivatives and I've never seen it calculated on "strange" boundaries like $\mathrm{\delta <!--\; \delta \; -->}\mathrm{\Omega <!--\; \Omega \; -->}$. Could someone briefly describe how we can derive (1)?

Could anyone help me with this question?
I need to compute the flux of vector field curl F through the hemisphere
$x=\sqrt{1-<!--\; -\; -->y^2-<!--\; -\; -->z^2}$
Positively oriented, with the vector field
$F(x,y,z)=⟨<!--\; ⟨\; -->e^{xy}cosz,x^{2}z,xy⟩<!--\; ⟩\; -->$

Given vectors (1,3,5),(−2,−6,−10) and (2,6,10) determine whether the linear span of the above is a plane in ${\mathbb{R}}^3$
The vectors are linearly dependent nd hence do not form a basis and it is known that the set of linearly dependent vectors in R^2 are collinear.
So based on the above can it be said that linearly dependent vectors in R^3 will form a plane.

The line is perpendicular to the plane 2x–3y+z–w=7 and passes the origin. Find parametric form vector form
This is my solution:
The line perpendicular to the plane implies being parallel to the normal vector. n=[2,−3,1,−1]
vector form: [x,y,z,w,v]=[0,0,0,0]+t[2,−3,1,−1]
parametric form: x=2t,y=−3t,z=t,w=−t,v=0
right?

How to show a curve lies in a plane given 3 constant vectors
How does one show that given 3 non-parallel vectors in ${\mathbb{R}}^3$, a curve with parameterisation:
$\mathbf{\text{s}}(t)=(x_1{t}^2+y_{1}t+c_1,x_2{t}^2+y_{2}t+c_2,x_3{t}^2+y_{3}t+c_3)$
lies in a plane?
Additionally, how do we find the equation of the plane?

Curvature defined as $k(t)=\frac{\Vert <!--\; \Vert \; -->T^{\prime }(t)\Vert <!--\; \Vert \; -->}{v(t)}$, explanation of the proof
I have problems to understand how it's derived the formula for the curvature $k(t)=\frac{\Vert <!--\; \Vert \; -->T^{\prime }(t)\Vert <!--\; \Vert \; -->}{v(t)}$
The derivation starts with the equality
$\frac{dT}{ds}\frac{ds}{dt}=\frac{dT}{dt}$
where T is the unit velocity vector for the curve and s is the arc length function for the curve.
So
$\frac{dT}{ds}=\frac{T^{\prime }(t)}{v(t)}$
and from here it's all clear to me how to finish.
Initially I supposed it was just an application of chain rule, but if I rewrite the proof in Netwon's notation I obtain
$[T(s(t)){]}^{\prime }=T^{\prime }(s(t))s^{\prime }(t)=T^{\prime }(s(t))v(t)$So to get the final equality I should have $[T(s(t)){]}^{\prime }=T^{\prime }(t)$, but is it always the case ?
I don't know if it is useful the fact that T(s(t)) and T(t) have the same trajectory (in general they differ for a certain value of t though). Is this enough? Why?

Let x,y and z be three linearly independent vectors. Explain, with justification, whether or not $\mathrm{span}<!--\; \; -->\{x,y,z\}=\mathrm{span}<!--\; \; -->\{x+y,y+z,z+x\}$

To ensure constant plane curvature, the curvature and unit binormal vector of the curve must possess constant values as given in the following.
I don't understand that the meaning of binormal vector is constant.
As far as I know, the binormal vector B is a vector vertical to osculating plane which is configured of the tangent vector, T, and normal vector N by $B=T\times <!--\; \times \; -->N$

I have
$A=\left(\begin{array}{cc}1& 2\\ 2& 2\\ 2& 1\end{array}\right)$
I have found that $range(A)=\{\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right),\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)\}$
and that $null(A^T)=\left\{\left(\begin{array}{c}1\\ -<!--\; -\; -->\frac{3}{2}\\ 1\end{array}\right)\right\}$
I have been asked to state the rank and nullity theorem for $A$ and $A^T$. I know the rank and nullity theorem as rank(A) + nullity(A) = n where A is an mxn matrix.
I know that nullity(A) is the number of vectors present in the null space of matrix A, but I'm not too sure how to go on from there.
Can someone help? The solution states dim(null($A^T$)) + dim(range($A$)) = 3

I'm wondering that why people doesn't care about vector's location. but, when we add the vectors, we move two or one vector to unite the vector's start point. and this mean location is important. but, when we learn very first of vector, we learn the vector doesn't matter where it is. in this point, I'm so confused because I can't find the reason that why vector's location doesn't matter. so i want to ask you - why the vector's location is not important?

Given $\stackrel{\to <!--\; \to \; -->}{u}=(\genfrac{}{}{0ex}{}{u_1}{u_2})$, if it is rotated sixty degrees anti-clockwise prove that the resulting vector will be $\stackrel{\to <!--\; \to \; -->}{v}=\frac{1}{2}(\genfrac{}{}{0ex}{}{u_1-<!--\; -\; -->\sqrt{3}{u}_2}{\sqrt{3}{u}_1+u_2})$
I think there are better methods out there but one idea I had was to use$\cos <!--\; \; -->60=\frac{1}{2}=\frac{\stackrel{\to <!--\; \to \; -->}{u}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{v}}{\left|\stackrel{\to <!--\; \to \; -->}{u}\right|\left|\stackrel{\to <!--\; \to \; -->}{v}\right|}$
and solve for $\stackrel{\to <!--\; \to \; -->}{v}$ but obviously this seems impossible to me with the $|\stackrel{\to <!--\; \to \; -->}{v}|$ and I see no where to force a $\sqrt{3}$ to appear.
Is this true? Or if this method works please share. I have other ideas which probably work but this one also came to mind but with no fruitition

Consider the space
${Ł}^p(\mathrm{\Omega <!--\; \Omega \; -->}):=(L^p(\mathrm{\Omega <!--\; \Omega \; -->}){)}^N=L^p(\mathrm{\Omega <!--\; \Omega \; -->})\times L^p(\mathrm{\Omega <!--\; \Omega \; -->})\times ...\times L^p(\mathrm{\Omega <!--\; \Omega \; -->}),N\ge <!--\; \ge \; -->1,\mathrm{\Omega <!--\; \Omega \; -->}\subset <!--\; \subset \; -->{\mathbb{R}}^{\mathbb{n}},$
as the vectorial $L^p$ space associated to the scalar one.
My questions are:
What is the norm $||\cdot <!--\; \cdot \; -->|{|}_{{Ł}^p}$ and can we relate it to $||\cdot <!--\; \cdot \; -->|{|}_{L^p}$?
How to apply the Hölder Inequality on
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}u(x)\cdot <!--\; \cdot \; -->v(x)dx,$
where
$u\in <!--\; \in \; -->{Ł}^q$ , $v\in <!--\; \in \; -->{Ł}^p$ , $\frac{1}{p}+\frac{1}{q}=1$, $u=(u_1,...,u_N{)}^t$, $v=(v_1,....,v_N{)}^t$ , $u_i\in <!--\; \in \; -->L^q(\mathrm{\Omega <!--\; \Omega \; -->})$, $v_i\in <!--\; \in \; -->L^p(\mathrm{\Omega <!--\; \Omega \; -->})$, and $u(x)\cdot <!--\; \cdot \; -->v(x)$ is the scalar product between?

What do these terms mean? $v(a)+R(h)$
In regards to the differentiability of a function at a.
I know that is it differentiable if the limit of $\frac{f(a+h)-<!--\; -\; -->f(a)}{h}$ exists.
Then I am given the following equation, supposing f is differentiable:
$f(a+h)-<!--\; -\; -->f(a)=h\cdot <!--\; \cdot \; -->v(a)+R(h)$ with $li{m}_{h\to <!--\; \to \; -->0}\frac{R(h)}{||h||}$
What is v(a) and R(h). Where did they come from?

Consider the vector $D=3\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ ar (unit vector).
Evaluate the integral
${\iint <!--\; \iint \; -->}_{S}DdS,$
where S is the surface of a sphere with radius r=5 centered at the origin.

Determine $\beta$ and $\alpha$ by using vectors such that A, B and C lie in the same plane, given that vector $\stackrel{\to <!--\; \to \; -->}{AB}=-<!--\; -\; -->4\stackrel{\to <!--\; \to \; -->}{\mathrm{ı<!--\; ı\; -->}}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{ȷ<!--\; ȷ\; -->}}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{k}$ and vector $\stackrel{\to <!--\; \to \; -->}{BC}=4\stackrel{\to <!--\; \to \; -->}{\mathrm{ı<!--\; ı\; -->}}+(\beta +3)\stackrel{\to <!--\; \to \; -->}{\mathrm{ȷ<!--\; ȷ\; -->}}+(\alpha -<!--\; -\; -->6)\stackrel{\to <!--\; \to \; -->}{k}$

If A is a $3\times <!--\; \times \; -->3$-matrix, while x and y are a vector point with x,y,z. Why is the dot product of $⟨<!--\; ⟨\; -->A^{\ast <!--\; \ast \; -->}x,y⟩<!--\; ⟩\; -->$ the same as the dot product of $⟨<!--\; ⟨\; -->A^{\prime \ast <!--\; \ast \; -->}y,x⟩<!--\; ⟩\; -->$
(A′ being the transposed matrix)

Suppose that I have a set of k values $s_1,\dots <!--\; \dots \; -->,s_k\in <!--\; \in \; -->K$ and k vectors $u_1,\dots <!--\; \dots \; -->,u_k\in <!--\; \in \; -->K^r$ for a field K not necessarily closed and k<r . Under which very general circumstances can I find a vector $w\in <!--\; \in \; -->K^r$ such that $w\cdot <!--\; \cdot \; -->u_k=s_k$ for each k? Observe that orthogonality would correspond to $s_k=0$ for all k.

Let $g:{\mathbb{R}}^{\mathbb{2}}\to <!--\; \to \; -->\mathbb{R}$ be totally differentiable in the point $a\in <!--\; \in \; -->{\mathbb{R}}^{\mathbb{2}}$
Let $v_1:=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)$ and $v_2:=\frac{1}{5}\left(\begin{array}{c}4\\ -<!--\; -\; -->3\end{array}\right)$ be normalized direction vectors in ${\mathbb{R}}^{\mathbb{2}}$
It is $\frac{\mathrm{\partial <!--\; \partial \; -->}g}{\mathrm{\partial <!--\; \partial \; -->}{v}_1}(a)=5\sqrt{2}$ and $\frac{\mathrm{\partial <!--\; \partial \; -->}g}{\mathrm{\partial <!--\; \partial \; -->}{v}_2}(a)=1$
With the info given above, how can one find out $\mathrm{\nabla <!--\; \nabla \; -->}g(a)$?

We know that, $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ satisfies $f^{\prime }(x)=f(x)$ for all $x\in <!--\; \in \; -->\mathbb{R}$ then $f(x)=A{e}^x$
What is a function $f:{\mathbb{R}}^d\to <!--\; \to \; -->{\mathbb{R}}^d$, analogous to the exponential function, if $\mathbf{x}\in <!--\; \in \; -->{\mathbb{R}}^d$ was a d dimensional vector?
Is there a name given to such functions whose derivatives are the same as the function itself?