Vector Examples, Equations, and Practice Problems

I am trying to write the vector $v=(2,3)$ in the basis outlined by the set $\{(-1,1),(3,-5)\}$. I figured that I would take the inverse of the set matrix, and multiply it to v. However, this does not seem to be the solution. Any help would be greatly appreciated.

Suppose $r(t)$ and $s(t)$ are vector functions with $r(2)=⟨<!--\; ⟨\; -->1,2,-1⟩<!--\; ⟩\; -->$, $r^{\prime }(t)=⟨<!--\; ⟨\; -->3,0,4⟩<!--\; ⟩\; -->$, and $s(t)=⟨<!--\; ⟨\; -->t,t^2,t^3⟩<!--\; ⟩\; -->$
(a) Find the value of f′(2), when $f(t)=r(t)·s(t)$
(b) Find the value of u′(2), when $u(t)=r(t)\times <!--\; \times \; -->s(t)$
To find the value of f(2) and u(2), I need r(t) which I do not know how to find. My guess is that since $r^{\prime }(t)=⟨<!--\; ⟨\; -->3,0,4⟩<!--\; ⟩\; -->$ then r(t) could equal $⟨<!--\; ⟨\; -->3x,0,4z⟩<!--\; ⟩\; -->$

$L_1$ is defined by ${\displaystyle \frac{x+1}{2}}={\displaystyle \frac{y-<!--\; -\; -->3}{3}}=1-<!--\; -\; -->z$
So we can write this as $((2t-<!--\; -\; -->1),(3t+3),(1-<!--\; -\; -->t))=r$
$L_2$ passes through (5,4,2) and intersects with $L_1$ at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z) So I got this 2x+3y−z=0 and said that x=3,y=2 and z=12
So now $L_2$ has the following equation r=((5+3t),(4+2t),(2+12t)) However when i equate $L_1$ and $L_2$ to find t. The equations aren't satisfied, meaning that t isn't common.

I can't figure out how to prove this using Index/Cartesian Tensor notation $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->(ab)=a\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}b+b(\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->a)$
Please note.....on the left side, that is NOT a dot product between vectors a and b. I'm assuming the problem wants them multiplied together to make a vector "$a_1{b}_1,a_2{b}_2,a_3{b}_3$".

How to derive $u^{T}u\cdot <!--\; \cdot \; -->v-<!--\; -\; -->v^{T}u\cdot <!--\; \cdot \; -->u$

How can one obtain (5.12) from (5.11)?
The relation
$\begin{array}{}\text{(5.11)}& (\mathbf{a}\times <!--\; \times \; -->\mathbf{b})\cdot <!--\; \cdot \; -->(\mathbf{c}\times <!--\; \times \; -->\mathbf{d})=(\mathbf{a}\cdot <!--\; \cdot \; -->\mathbf{c})(\mathbf{b}\cdot <!--\; \cdot \; -->\mathbf{d})-<!--\; -\; -->(\mathbf{a}\cdot <!--\; \cdot \; -->\mathbf{d})(\mathbf{b}\cdot <!--\; \cdot \; -->\mathbf{c})\end{array}$
is called the Identity of Langrange. From (5.11) follows
$\begin{array}{}\text{(5.12)}& (\mathbf{a}\times <!--\; \times \; -->\mathbf{b})\times <!--\; \times \; -->\mathbf{c}=(\mathbf{a}\cdot <!--\; \cdot \; -->\mathbf{c})\mathbf{b}-<!--\; -\; -->(\mathbf{b}\cdot <!--\; \cdot \; -->\mathbf{c})\mathbf{a}\end{array}$
Can someone give a hint?

How could you find a vector if you know the values of another vector and the angle between them in three dimensions? Also, the y component is the same for both vectors and the magnitude of both vectors is equal to 1.

The question says
The volume of a tetrahedron DABC is 9 cubic units. If $\mathrm{\angle <!--\; \angle \; -->}ACB=\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$ and $2AD+AC+BC=18$, then the length AD is
What I could come up with
If I set up a coordinate system with D at origin, and other points with their respective vectors
$9=\frac{1}{6}(\stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{b}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{c}))$
To solve for AD, solving for AC+BC suffices. Cosine rule combined with dot products only leads me to
$(AC+BC{)}^2=A{B}^2+(2+\sqrt{3})AC\cdot <!--\; \cdot \; -->BC={|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}|}^2+\frac{(4+2\sqrt{3})}{\sqrt{3}}((\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a})\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{b}))$
I dont see how this connected to the scalar triple product, nor do I know if this is the correct start/approach to this problem.

For parallel translation $T_v$, does $T_{-<!--\; -\; -->v}={T_v}^{-<!--\; -\; -->1}$ hold true?

$f_{\mathbf{X}}(x_1,\dots <!--\; \dots \; -->,x_k)=\frac{\exp <!--\; \; -->(-<!--\; -\; -->\frac{1}{2}(\mathbf{x}-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}{)}^{\mathrm{T}}{\mathbf{\Sigma <!--\; \Sigma \; -->}}^{-<!--\; -\; -->1}(\mathbf{x}-<!--\; -\; -->\mathit{\mu <!--\; \mu \; -->}))}{\sqrt{(2\mathrm{\pi <!--\; \pi \; -->}{)}^k|\mathbf{\Sigma <!--\; \Sigma \; -->}|}}$
I believe $x_1,...x_k\in <!--\; \in \; -->{\mathbb{R}}^d$
How is range of $f_{\mathbf{X}}$ a scalar?

Find the unit vector $\stackrel{\to <!--\; \to \; -->}{b}$ if $-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{k}$ bisects the angle between $\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{a}=3\stackrel{^<!--\; ^\; -->}{i}+4\stackrel{^<!--\; ^\; -->}{j}$
The unit vector along the angle bisector is
$\frac{-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{k}}{\sqrt{3}}=\frac{\stackrel{\to <!--\; \to \; -->}{b}+(\frac{3}{5}\stackrel{^<!--\; ^\; -->}{i}+\frac{4}{5}\stackrel{^<!--\; ^\; -->}{j})}{|\stackrel{\to <!--\; \to \; -->}{b}+\frac{3}{5}\stackrel{^<!--\; ^\; -->}{i}+\frac{4}{5}\stackrel{^<!--\; ^\; -->}{j}|}$
Im not able to extract $\stackrel{\to <!--\; \to \; -->}{b}$ from here

Can the gradient exist for a function of n+1 variables?
For a function of $n+1$ variables $f(x_0,x_1,x_2,...x_n)$ can a gradient exist?
When I asked my professor this during class he said, "no, at most a gradient will exist for a function of three variables f(x,y,z) because there are only at most three standard basis vectors with which to represent a vector."
This is a calculus 3 class so perhaps this answer was given to keep the concept of the gradient within the scope of the class, but I suspect this isn't the whole story and there is more to this than my professor is telling.
Edit:
The definition of the gradient for a function of two variables given during class was: Let z=f(x,y) be a function, then the gradient of f is defined as $\mathrm{\nabla <!--\; \nabla \; -->}f=f_x\stackrel{\to <!--\; \to \; -->}{i}+f_y\stackrel{\to <!--\; \to \; -->}{j}$

In $\mathrm{\Delta <!--\; \Delta \; -->}ABC$, $\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=2$ find circumcentre given coordinates of P,Q
Let $A,B,C$ be the vertices of $\mathrm{\Delta <!--\; \Delta \; -->}ABC$. Let $P(1,1,2)$ and $Q(4,7,8)$ be two points such that $\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=2$. Find the coordinates of circumcentre of $\mathrm{\Delta <!--\; \Delta \; -->}ABC$

Find vectors u,v that satisfy the equation $Au=e_1$, and $Av=e_2$
I am really new to matrices and these concepts, so any help would be much appreciated.
$A=\left[\begin{array}{ccc}1& -<!--\; -\; -->3& 0\\ 1& 4& -<!--\; -\; -->3\\ 0& 1& 5\end{array}\right]$
We're asked to let each row = $a_1,a_2,a_3$ going down and then calculate the cross products and triple scalar products which I was able to do.
I calculated the cross products of $a_1\times <!--\; \times \; -->a_3$, and then $a_2\times <!--\; \times \; -->a_3$: which are:
$a_1\times <!--\; \times \; -->a_3=\left[\begin{array}{c}-<!--\; -\; -->15\\ -<!--\; -\; -->5\\ 1\end{array}\right]$
and
$a_2\times <!--\; \times \; -->a_3=\left[\begin{array}{c}23\\ -<!--\; -\; -->5\\ 1\end{array}\right]$
If $Au=e_1$ and $Av=e_2$, and I need to determine vectors u and v, could someone explain how to find u and v. I know that if I multiply A by the cross product vectors I get [38,0,0] and [0,−38,0] but this confuses me because I thought to satisfy the equation the answer should be [1,0,0] and [0,1,0]. I have also tried finding the inverse of A and multiply this by $e_1$ and $e_2$, but the results were not even close to the result of the cross products?? I am really trying to learn, would appreciate some advice.
Is there another method to solve this?

I previously had to prove the statement, and I did so by showing there were scalars a and b that were equal to each other but now I have to give a counterexample. I just proved this hypothesis' correctness, did I prove it wrong or is there a proper counterexample?
For reference, I proved the hypothesis like this: Given vector w $\in <!--\; \in \; -->$ Span($v_1,v_2,v_3$), we have $a_1{v}_1+a_2{v}_2+a_3{v}_3$=w and $b_1{v}_1+b_2{v}_2+b_3{c}_3$ = w With some basic algebra, we can reach the point where $a_1-<!--\; -\; -->b_1=a_2-<!--\; -\; -->b_2=a_3-<!--\; -\; -->b_3=0$

Given two vectors $\stackrel{\to <!--\; \to \; -->}{a}=(5,7,-<!--\; -\; -->3)$ and $\stackrel{\to <!--\; \to \; -->}{b}=(1,0,-<!--\; -\; -->3)$
Find the projection of $\stackrel{\to <!--\; \to \; -->}{a}$ onto $\stackrel{\to <!--\; \to \; -->}{b}$
So we first find $|\stackrel{\to <!--\; \to \; -->}{a}|=\sqrt{83}$ and then $|\stackrel{\to <!--\; \to \; -->}{b}|=\sqrt{10}$
We find $\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ using
$\stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}=|\stackrel{\to <!--\; \to \; -->}{a}||\stackrel{\to <!--\; \to \; -->}{b}|cos\mathrm{\theta <!--\; \theta \; -->}$
Which is:
$14=\sqrt{83}\ast <!--\; \ast \; -->\sqrt{10}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}⟺<!--\; ⟺\; -->\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{14}{\sqrt{830}}$
Now we want to find $\stackrel{\to <!--\; \to \; -->}{a}\stackrel{^<!--\; ^\; -->}{b}$ where $\stackrel{^<!--\; ^\; -->}{b}=\frac{\stackrel{\to <!--\; \to \; -->}{b}}{|\stackrel{\to <!--\; \to \; -->}{b}|}$ so we get:
$\frac{\stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}}{|\stackrel{\to <!--\; \to \; -->}{b}|}=\frac{14}{\sqrt{10}}$
but it should have a direction, so the answer is $\frac{14}{\sqrt{10}}\stackrel{^<!--\; ^\; -->}{b}$? but we have used $\stackrel{^<!--\; ^\; -->}{b}=\frac{\stackrel{\to <!--\; \to \; -->}{b}}{|\stackrel{\to <!--\; \to \; -->}{b}|}$ in the calculation

Will a vector rotate back to itself under unitary rotation?
Let there be an arbitrary complex vector $\stackrel{\to <!--\; \to \; -->}{v}\in <!--\; \in \; -->{\mathbb{C}}^D$ that has unit length defined as $\underset{i}{\sum <!--\; \sum \; -->}{v}_i{v}_i^{\ast <!--\; \ast \; -->}=1$. Let there be a Hermitian $D\times <!--\; \times \; -->D$ matrix H. Given arbitrary $\stackrel{\to <!--\; \to \; -->}{v}$ and H, will there always exist a non-zero $t\in <!--\; \in \; -->\mathbb{R}$ such that $e^{iHt}\stackrel{\to <!--\; \to \; -->}{v}=\stackrel{\to <!--\; \to \; -->}{v}$?

If I want to find a function $V(r)$ such that $\mathbf{F}=-<!--\; -\; -->\mathrm{\nabla <!--\; \nabla \; -->}V$, where $\mathbf{F}=f(r)\stackrel{^<!--\; ^\; -->}{\mathbf{r}}$, how do I integrate F along the radial direction?

I am studying arclength for a path c(t)=(x(t),y(t),z(t)). I understand that we integrate the derivative of speed etc.
My question is, geometrically, what is |c(t)|? c(t) is a formula of displacement, so my naive interpretation is that taking its magnitude would yield distance travelled, but this clearly isn't the case.
As a follow-up, what is the difference between the rate of change of displacement and the rate of change of distance?

Can someone check my math here? I feel like this should be a very simple problem, but I can't seem to confirm this by searching.
What is the derivative of a vector norm? I'm referring to the usual Euclidean norm (2-norm)
$s=\Vert <!--\; \Vert \; -->\mathbf{x}\Vert <!--\; \Vert \; -->\text{ where }\mathbf{x}\in <!--\; \in \; -->{\mathbb{R}}^3\text{ and }\mathbf{x}=f(t)$
Is it correct that:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{s}=\frac{\mathbf{x}\cdot <!--\; \cdot \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathbf{x}}}{\Vert <!--\; \Vert \; -->\mathbf{x}\Vert <!--\; \Vert \; -->}$
The dots are referring to time derivatives $(\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathbf{x}}=\frac{d}{dt}\mathbf{x})$
If the above is not correct, how to I calculate $\stackrel{\dot{}<!--\; \dot{}\; -->}{s}$ from $\mathbf{x}$ and $\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathbf{x}}$