Vector Examples, Equations, and Practice Problems

The vector $A=5i+6j$ is rotated through an $\mathrm{\angle <!--\; \angle \; -->}45$ about the Z axis in the anticlockwise direction. What is the resultant vector?
My attempt: I tried to calculate the resultant vector by using the equation, $R=\sqrt{A^2+B^2+2ABCos\mathrm{\theta <!--\; \theta \; -->}}$
since it is rotated in anticlockwise direction its direction changes. Any hint will be appreciated.

If $\stackrel{\to <!--\; \to \; -->}{a},\stackrel{\to <!--\; \to \; -->}{b}\mathrm{\&<!--\; \&\; -->}\stackrel{\to <!--\; \to \; -->}{c}$ are unit vector such that ${\left|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}\right|}^2=9$
Then the value $\left|2\stackrel{\to <!--\; \to \; -->}{a}+5\stackrel{\to <!--\; \to \; -->}{b}+5\stackrel{\to <!--\; \to \; -->}{c}\right|=\mathrm{\_<!--\; \_\; -->}\mathrm{\_<!--\; \_\; -->}\mathrm{\_<!--\; \_\; -->}\mathrm{\_<!--\; \_\; -->}\mathrm{\_<!--\; \_\; -->}\mathrm{\_<!--\; \_\; -->}$
My approach is as follow
${\left|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}\right|}^2={\left|\stackrel{\to <!--\; \to \; -->}{a}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{b}\right|}^2-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{a}.\stackrel{\to <!--\; \to \; -->}{b}+{\left|\stackrel{\to <!--\; \to \; -->}{b}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{c}\right|}^2-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{b}.\stackrel{\to <!--\; \to \; -->}{c}+{\left|\stackrel{\to <!--\; \to \; -->}{c}\right|}^2+{\left|\stackrel{\to <!--\; \to \; -->}{a}\right|}^2-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{c}.\stackrel{\to <!--\; \to \; -->}{a}=9$
$\left(\stackrel{\to <!--\; \to \; -->}{a}.\stackrel{\to <!--\; \to \; -->}{b}+\stackrel{\to <!--\; \to \; -->}{b}.\stackrel{\to <!--\; \to \; -->}{c}+\stackrel{\to <!--\; \to \; -->}{c}.\stackrel{\to <!--\; \to \; -->}{a}\right)=-<!--\; -\; -->\frac{3}{2}$
How do I proceed from here

What is the difference between $\frac{(\stackrel{\to <!--\; \to \; -->}{B}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->})\stackrel{\to <!--\; \to \; -->}{B}}{B}$ and $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B}$?
I was doing a vector calculus problem, and in the hint I got to introduced with the first term, and I wondered, what is the difference between these two expression?

Is (2,0,1) an eigenvector of A?
Let A = $\left(\begin{array}{ccc}5& 1& 2\\ 0& 6& 0\\ 1& -<!--\; -\; -->1& 4\end{array}\right)$
(**) $A\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}3\\ 0\\ -<!--\; -\; -->3\end{array}\right)$ , $A\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}6\\ 6\\ 0\end{array}\right)$, $A\left(\begin{array}{c}1\\ -<!--\; -\; -->1\\ 1\end{array}\right)=\left(\begin{array}{c}6\\ -<!--\; -\; -->6\\ 6\end{array}\right)$, $\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+\left(\begin{array}{c}1\\ -<!--\; -\; -->1\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)$
Which of the following is true?
(1) (2,0,1) is an eigenvector associated to 3
(2) (2,0,1) is an eigenvector associated to 6
(3) (2,0,1) is not an eigenvector of A
I found that only (2) is true by using the characteristic polynomial, but how do I prove that (2) is true using (**), as intended by the question?

I have a feeling this is a very easy question but I can't figure out what I'm missing
Let's say we have a quadratic form
$f(x)=\frac{1}{2}{x}^{T}Ax-<!--\; -\; -->b^{T}x$
where A is a $n\times <!--\; \times \; -->n$ symmetric and positive definite matrix and x,b are $n\times <!--\; \times \; -->1$ vectors.
If we introduce the weighted norm as
$\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_A^2=x^{T}Ax$
and x∗ is the solution of the system Ax=b, then I want to prove that
$\frac{1}{2}\Vert <!--\; \Vert \; -->x-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}{\Vert <!--\; \Vert \; -->}_A^2=f(x)-<!--\; -\; -->f(x^{\ast <!--\; \ast \; -->})$
What I've done so far:
$$\begin{gathered} \frac{1}{2}\Vert <!--\; \Vert \; -->x-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}{\Vert <!--\; \Vert \; -->}_A^2 \\ =\frac{1}{2}(x-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}{)}^{T}A(x-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}) \\ =\frac{1}{2}(x^{T}Ax-<!--\; -\; -->x^{T}A{x}^{\ast <!--\; \ast \; -->}-<!--\; -\; -->x^{\ast <!--\; \ast \; -->T}Ax+x^{\ast <!--\; \ast \; -->T}A{x}^{\ast <!--\; \ast \; -->}) \\ =\frac{1}{2}(x^{T}Ax-<!--\; -\; -->b^{T}x-<!--\; -\; -->b^{T}x+x^{\ast <!--\; \ast \; -->T}A{x}^{\ast <!--\; \ast \; -->}) \\ =\frac{1}{2}{x}^{T}Ax-<!--\; -\; -->b^{T}x+\frac{1}{2}{x}^{\ast <!--\; \ast \; -->T}Ax= \\ f(x)+\frac{1}{2}{x}^{\ast <!--\; \ast \; -->T}A{x}^{\ast <!--\; \ast \; -->} \end{gathered}$$
I'm missing a term $-<!--\; -\; -->b^T{x}^{\ast <!--\; \ast \; -->}$ to complete the proof but no matter how I look at it I can't find where I've made a mistake, if there is one.

why a vector field
$\mathbf{w}(x,y,z)=f_1{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_x+f_2{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_y+f_3{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_z$
can be associated with the 2-form
${\mathrm{\omega <!--\; \omega \; -->}}_2=f_{1}dy\wedge <!--\; \wedge \; -->dz+f_{2}dz\wedge <!--\; \wedge \; -->dx+f_{3}dx\wedge <!--\; \wedge \; -->dy.$
Is it OK to say this works because the wedge product of the 1-forms $\mathrm{\omega <!--\; \omega \; -->}=u^{1}dx+u^{2}dy+u^{3}dz$ and $\mathrm{\nu <!--\; \nu \; -->}=v^{1}dx+v^{2}dy+v^{3}dz:$
$\mathrm{\omega <!--\; \omega \; -->}\wedge <!--\; \wedge \; -->\mathrm{\nu <!--\; \nu \; -->}=(u^2{v}^3-<!--\; -\; -->u^3{v}^2)dy\wedge <!--\; \wedge \; -->dz+(u^3{v}^1-<!--\; -\; -->u^1{v}^3)dz\wedge <!--\; \wedge \; -->dx+(u^1{v}^2-<!--\; -\; -->u^2{v}^1)dx\wedge <!--\; \wedge \; -->dy$
has the same components as the cross product of the vectors $\mathbf{u}=u^1{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_x+u^2{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_y+u^3{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_z$ and $\mathbf{v}=v^1{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_x+v^2{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_y+v^3{\stackrel{^<!--\; ^\; -->}{\mathbf{e}}}_z$?

Reparameterize the curve $r(t)=⟨<!--\; ⟨\; -->e^t\sin <!--\; \; -->t,e^t\cos <!--\; \; -->t,5e^t⟩<!--\; ⟩\; -->$ in terms of the arclength parameter, s with (0,1,5) as the base point.
So first, $r^{\mathrm{\prime <!--\; \prime \; -->}}(t)=⟨<!--\; ⟨\; -->e^t\sin <!--\; \; -->t+e^t\cos <!--\; \; -->t,e^t\cos <!--\; \; -->t-<!--\; -\; -->e^t\sin <!--\; \; -->t,5e^t⟩<!--\; ⟩\; -->$
Then I took the magnitude of r′(t) which is $\sqrt{(e^t\sin <!--\; \; -->t+e^t\cos <!--\; \; -->t{)}^2+(e^t\cos <!--\; \; -->t-<!--\; -\; -->e^t\sin <!--\; \; -->t{)}^2+(5e^t{)}^2}$
Next, I took the integral from 0 to t of $\sqrt{(e^t\sin <!--\; \; -->t+e^t\cos <!--\; \; -->t{)}^2+(e^t\cos <!--\; \; -->t-<!--\; -\; -->e^t\sin <!--\; \; -->t{)}^2+(5e^t{)}^2}dt$ to get $s(t)=3\sqrt{3}{e}^t$
After this I don't know how to get t(s) and then r(t(s)).

Given matrices A,B,C and column vector $\mathbf{\text{v}}$, what is the derivative of $⟨<!--\; ⟨\; -->\mathbf{\text{A}}\text{diag}(\mathbf{\text{B}}\mathbf{\text{v}}),\mathbf{\text{C}}⟩<!--\; ⟩\; -->$ with respect to $\mathbf{\text{B}}$? diag(⋅) is a diagonal matrix with the argument as the diagonal and the brackets signify the inner product.
I am trying to find the solution, but am hung up on differentiating the diagonal matrix.

Suppose I have a set of vectors $X=\{x_i{\}}_{i=1}^n\subset <!--\; \subset \; -->{\mathbb{R}}^d$. Fix a vector $x\in <!--\; \in \; -->X$ and assume that a different vector y is one of the K points in X that is closest to x in Euclidean distance. In other words, y is one of the K nearest neighbors of x. Does this mean that x is one of the K nearest neighbors of y too?
Basically, is the statement "x is one of the K nearest neighbors for y" reflexive?

I would like to integrate
$\int <!--\; \int \; -->d^3\stackrel{\to <!--\; \to \; -->}{k}\int <!--\; \int \; -->d^3\stackrel{\to <!--\; \to \; -->}{q}\frac{1}{q^2}$
under the condition that
$|\stackrel{\to <!--\; \to \; -->}{k}|\le <!--\; \le \; -->A\text{and}|\stackrel{\to <!--\; \to \; -->}{k}+\stackrel{\to <!--\; \to \; -->}{q}|\le <!--\; \le \; -->A$
where A is some constant.
Is this executable?

Prove that the vectors i−k, −i+j+2k, 2i−j−3k can form the sides of the triangle.
I tried adding these vectors together, with the aim of getting a zero vector but I ended up with the following:
$\mathbf{i}(1-<!--\; -\; -->1+2)+\mathbf{j}(0+1-<!--\; -\; -->1)+\mathbf{k}(-<!--\; -\; -->1+2-<!--\; -\; -->3)=2\mathbf{i}-<!--\; -\; -->2\mathbf{k}$
Rather than first informing my tutor about a possible mistake, I am first trying to check whether I am doing something wrong from my side.

I know the vector identity $\stackrel{\to <!--\; \to \; -->}{A}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{B}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{C})=(\stackrel{\to <!--\; \to \; -->}{A}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{C})\stackrel{\to <!--\; \to \; -->}{B}-<!--\; -\; -->(\stackrel{\to <!--\; \to \; -->}{A}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B})\stackrel{\to <!--\; \to \; -->}{C}$
Now, is there a succinct way of obtaining $|\stackrel{\to <!--\; \to \; -->}{A}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{B}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{C}){|}^2$ using vector algebra? I know we can expand, multiply and group the terms back, but is there a neater way of obtaining the result?
Using Levi-Civita symbols, would this be easier?

Find two vectors u and v such that: (i) both u and v are perpendicular to $w=(1,-<!--\; -\; -->1,-<!--\; -\; -->1)$ and (ii) u is perpendicular to v
So, I know how to find a vector, $u=(1,2,-<!--\; -\; -->1)$, that is perpendicular to w, but how do I find the third vector, v that is perpendicular to both u and w?
I've already tried
$v\cdot <!--\; \cdot \; -->(u+w)=v\cdot <!--\; \cdot \; -->u+v\cdot <!--\; \cdot \; -->w$
But the result that I get does not allow for $u\cdot <!--\; \cdot \; -->v=0$ and $v\cdot <!--\; \cdot \; -->w=0$
What am I doing wrong? How should I approach solving something like this?

An object of mass 10 moves with position function $r(t)=<1+3\cos <!--\; \; -->(t),4\cos <!--\; \; -->(t),-1+5\sin <!--\; \; -->(t)>$ Find the force vector acting on this object at time t. Show that it points from the object towards the point $(1,0,-1)$
I believe that I have found the correct force vector: $F(t)=10<-<!--\; -\; -->3\cos <!--\; \; -->t,-<!--\; -\; -->4\cos <!--\; \; -->t,-<!--\; -\; -->5\sin <!--\; \; -->t>$
However, I am not sure how to figure out how to show that it points from the object towards the point $(1,0,-<!--\; -\; -->1)$
Any help would be greatly appreciated.

How is the following decomposition done?
$\begin{array}{rl}\left[\begin{array}{cc}1-<!--\; -\; -->a& a\\ b& 1-<!--\; -\; -->b\end{array}\right]& =\frac{1}{a+b}\left[\begin{array}{cc}b& a\\ b& a\end{array}\right]+\frac{1-<!--\; -\; -->a-<!--\; -\; -->b}{a+b}\left[\begin{array}{cc}a& -<!--\; -\; -->a\\ -<!--\; -\; -->b& b\end{array}\right].\end{array}$
How is the above decomposition done?
Can anyone explain?

Is this a vector space and is this the correct way to approach the problem?
See if, based on the operations presented, it is a vector space in the real numbers.
For any
$(x_1,x_2),(y_1,y_2)\in <!--\; \in \; -->{\mathbb{R}}^2;a\in <!--\; \in \; -->\mathbb{R}$
The operations presented are:
$(x_1,x_2)+(y_1,y_2)=(x_1,x_2-<!--\; -\; -->y_2)$
and
$a(x_1,x_2)=(a{x}_1,a{x}_2)$
My first question is if, based on them saying
a(x1,x2)=(ax1,ax2)
does it mean that
$a(x_1,x_2)=(a{x}_1,a{x}_2)$?
and would this apply if the multiplication was defined in any other way, like:
$a(x_1,x_2)=(0,a{x}_1)$
then we could assume that
$a(y_1,y_2)=(0,a{y}_1)?$
I tried to verify the first axiom based on their definition and my thought was as follows:
$(x_1,x_2)+(y_1,y_2)=(x_1,x_2-<!--\; -\; -->y_2)$
$(y_1,y_2)+(x_1,x_2)=(y_1+x_1,y_2+x_2)$
This means that it doesn't fit the criteria to be a vector space... Is this the correct way to prove it? Also, I'm not sure if the order of the elements in the sums is relevant or not... I'm assuming it is based on their definition, but since we are dealing with elements in the real numbers I could also assume they aren't. In that case we could get something like:
$(y_1,y_2)+(x_1,x_2)=(y_1+x_1,y_2+x_2)=(x_1+y_1,x_2+y_2)=(x_1,x_2-<!--\; -\; -->y_2)$
In this case, we have commutation and the first axiom holds.

Let u,v and w be vectors in ${\mathbb{R}}^n$ and let $\mathrm{\theta <!--\; \theta \; -->}(u,w),\mathrm{\theta <!--\; \theta \; -->}(v,w)$ and $\mathrm{\theta <!--\; \theta \; -->}(u+v,w)$ represent the angle between each listed pair of vectors. Does it hold that one of the following two statements must be true:
$\mathrm{\theta <!--\; \theta \; -->}(u,w)\ge <!--\; \ge \; -->\mathrm{\theta <!--\; \theta \; -->}(u+v,w)$
Or:
$\mathrm{\theta <!--\; \theta \; -->}(v,w)\ge <!--\; \ge \; -->\mathrm{\theta <!--\; \theta \; -->}(u+v,w)$
I feel this must be true and that it follows from some simple property that I'm forgetting. If it is true, does it generalize to any finite list of vectors?

If a vector $2i+3j+8k$ is perpendicular to the vector $4j-<!--\; -\; -->4i+\alpha k$ then the value of α is? solve without using the property of dot product.

So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis $\stackrel{\to <!--\; \to \; -->}{a}$,$\stackrel{\to <!--\; \to \; -->}{b}$ , and i want to find out the angle between its diagonals starting only from the fact that
$\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{a}\Vert <!--\; \Vert \; -->=3$
and
$\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{b}\Vert <!--\; \Vert \; -->=2$
and the angle between them pi/3 , how would i go about it?
I know the diagonals are $\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}$ respectively, but im not sure how I shoul apply the formula for sum and difference of vectors.

What does it mean to differentiate a vector?
I didn’t quite understand what does it mean to differentiate a vector, i suspect that the derivative of a vector valued function is just answering the question : What vector should i added to the previous one to get the next one. For example :
Consider the vector
$\stackrel{\to <!--\; \to \; -->}{X}(t)=⟨<!--\; ⟨\; -->1,t⟩<!--\; ⟩\; -->$
${\stackrel{\to <!--\; \to \; -->}{X}}^{\prime }(t)=⟨<!--\; ⟨\; -->0,1⟩<!--\; ⟩\; -->$
Now if i take a random vector from the vector valued function ⟨1,t⟩ , say for example ⟨1,3⟩ If i add to it the derivative of ⟨1,t⟩ which is ⟨0,1⟩, i get ⟨1,4⟩ which is the next vector.
But this is only true for $t\in <!--\; \in \; -->\mathbb{Z}$ because in the real numbers the next vector is not defined, it could be $⟨<!--\; ⟨\; -->1,3.0001⟩<!--\; ⟩\; -->$
So how can i really understand what is the derivative of a vector.