Vector Examples, Equations, and Practice Problems

Vectors a, b, c make ${60}^{\circ <!--\; \circ \; -->}$ angles with each other. $|a|=4$, $|b|=2$ , $|c|=6$. Find the length of p=a+b+c.
The only way I can think of a, b and c having ${60}^{\circ <!--\; \circ \; -->}$ angles with each other is that they form a vertex of a tetrahedron. Then, I can find |a+b| or |b+c| or |a+c| using the law of cosines. But then I can't find |p|, because I don't know the angle between the vector I have found and the remaining one.
I would like to get some hints or clues how to solve this, thanks in advance.

How to find the third point at a given distance angle from a line segment in a plane?
Suppose i have a plane and a line segment with end points $(x_1,y_1),(x_2,y_2)$ is given to us.Now we need to find another point which lies at distance d from $(x_2,y_2)$ , making an angle θ in such a way that if we go from $(x_1,y_1)$ to $(x_2,y_2)$ and then to $(x_3,y_3)$ it forms counterclockwise direction.
Note : $p1=(x_1,y_1),p2=(x_2,y_2),p3=(x_3,y_3)$
My approach : we can use dot product and it will give us equation of for Ax+By=C where A,B,C are constants.Then we can use information given by d. After solving two equations we will get two points , one in clockwise direction and other in counterclockwise direction . We can get out desired point using cross product (it will be negative for the point we want and positive for the other point).
You see that my approach is very lengthy .Is there any shortapproach ?

Suppose $a,b\in <!--\; \in \; -->{\mathbb{R}}^n$. Why is it the case that $(a^{T}b)b=(b{b}^T)a$?
I know that $a^{T}b=b^{T}a$, but I'm not sure why the above holds.

Let $T(x,y)=(x+2y,x-<!--\; -\; -->y)$ and $U(x,y)=(x-<!--\; -\; -->y,3x+y)$ be linear transformations from $R^2$ to $R^2$ Determine the transformations of the composition of U with $T$ $(U°T)$
So I can determine the standard matrices are $T=$$\left[\begin{array}{cc}1& 2\\ 1& -<!--\; -\; -->1\end{array}\right]$ and $U=$ $\left[\begin{array}{cc}1& -<!--\; -\; -->1\\ 3& 1\end{array}\right]$ but I'm not sure how to use these with compositions. What is the process needed to answer the question?

How to solve $k_b\stackrel{¯<!--\; ¯\; -->}{b}-<!--\; -\; -->k_a\stackrel{¯<!--\; ¯\; -->}{a}+\frac{1}{t}\cdot <!--\; \cdot \; -->k_a\stackrel{¯<!--\; ¯\; -->}{a}-<!--\; -\; -->\frac{1}{t}(\stackrel{¯<!--\; ¯\; -->}{c}-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{a}}_1)={\stackrel{¯<!--\; ¯\; -->}{a}}_1-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{b}}_1$

I have seen that the volume of a tetrahedron ABCD where
$A=(x_1,y_1,z_1),B=(x_2,y_2,z_2),C=(x_3,y_3,z_3),D=(x_4,y_4,z_4)$
and A being the top can be described as
$V={\displaystyle \frac{1}{6}}|detM|\text{where}M=\left[\begin{array}{cccc}{x}_1& y_1& z_1& 1\\ x_2& y_2& z_2& 1\\ x_3& y_3& z_3& 1\\ x_4& y_4& z_4& 1\end{array}\right]$
but I cannot find the proof anywhere. How is the following true?
$\stackrel{\to <!--\; \to \; -->}{BC}\cdot <!--\; \cdot \; -->(\stackrel{\to <!--\; \to \; -->}{BD}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{BA})=detM$
Any kind of explanation is welcome.

I have the following identity to prove:
$\mathrm{grad}<!--\; \; -->\mathrm{div}<!--\; \; -->\mathbf{\text{A}}=\mathrm{rot}<!--\; \; -->\mathrm{rot}<!--\; \; -->\mathbf{\text{A}}+\mathrm{div}<!--\; \; -->\mathrm{grad}<!--\; \; -->\mathbf{A}$
The part with gradA is confusing to me as A is a vector so how do we calculate gradient of it? Could it be that my professor formulated the identity wrong?

Let $f:{\mathbb{R}}^n\to <!--\; \to \; -->{\mathbb{R}}^n$ be a map where $\Vert <!--\; \Vert \; -->f(a)-<!--\; -\; -->f(b)\Vert <!--\; \Vert \; -->=\Vert <!--\; \Vert \; -->a-<!--\; -\; -->b\Vert <!--\; \Vert \; -->$ and f(0)=0, prove $a·b=f(a)·f(b)$
Obviously this is a linear transformation, but how do we go about proving this statement?

If $\mathbf{v}\cdot <!--\; \cdot \; -->\mathbf{A}=\mathbf{v}\cdot <!--\; \cdot \; -->\mathbf{B}$ for all $\mathbf{v}$, does it imply $\mathbf{A}=\mathbf{B}$? Can there be a simple proof for this?

What am I doing wrong here?
I am trying to show $\mid <!--\; \mid \; -->a-<!--\; -\; -->x|>|b-<!--\; -\; -->x|$ using the fact that $(a-<!--\; -\; -->b)\cdot <!--\; \cdot \; -->(x-<!--\; -\; -->b)<0$ for $a,b,x\in <!--\; \in \; -->{\mathbb{R}}^N$
So far, I have done this:
$(a-<!--\; -\; -->b)\cdot <!--\; \cdot \; -->(x-<!--\; -\; -->b)<0⟹<!--\; ⟹\; -->(a-<!--\; -\; -->b)\cdot <!--\; \cdot \; -->(b-<!--\; -\; -->x)>0$
$⟹<!--\; ⟹\; -->\text{by Cauchy-Schwarz}\mid <!--\; \mid \; -->a-<!--\; -\; -->b\mid <!--\; \mid \; -->\mid <!--\; \mid \; -->b-<!--\; -\; -->x\mid <!--\; \mid \; -->>0$
$⟹<!--\; ⟹\; -->\mid <!--\; \mid \; -->a-<!--\; -\; -->x+x-<!--\; -\; -->b\mid <!--\; \mid \; -->\mid <!--\; \mid \; -->b-<!--\; -\; -->x\mid <!--\; \mid \; -->>0$
$⟹<!--\; ⟹\; -->\text{using triangle inequality}(\mid <!--\; \mid \; -->a-<!--\; -\; -->x\mid <!--\; \mid \; -->+\mid <!--\; \mid \; -->x-<!--\; -\; -->b\mid <!--\; \mid \; -->)\mid <!--\; \mid \; -->b-<!--\; -\; -->x\mid <!--\; \mid \; -->>0$
$⟹<!--\; ⟹\; -->\mid <!--\; \mid \; -->a-<!--\; -\; -->x\mid <!--\; \mid \; -->>-<!--\; -\; -->\mid <!--\; \mid \; -->b-<!--\; -\; -->x\mid <!--\; \mid \; -->$
I shouldn't have the negative sign, but can't see where I am going wrong. Some advice would be appreciated. Thank you.

I have two 2D unit vectors a and b. I'd like to find the rotation matrix that rotates a to b. The formulas I see online are for a rotation matrix are
$\left(\begin{array}{cc}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}& -<!--\; -\; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\\ \sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}& \cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\end{array}\right)$
And I can get the angle between a and b with
$\mathrm{\theta <!--\; \theta \; -->}={\cos }^{-<!--\; -\; -->1}<!--\; \; -->(\mathbf{a}\cdot <!--\; \cdot \; -->\mathbf{b})$
My problem is that that doesn't give me the direction. For example, if $\mathrm{\theta <!--\; \theta \; -->}$ is $\mathrm{\pi <!--\; \pi \; -->}/2$ when maybe the matrix should use $-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}/2$

Is there a vector that forms angles of 30 and 45 degrees with the x and y axes respectively?
I've a problem that asks
Is there any 3D-vector that forms angles of 30 and 45 degrees with x-axis and y-axis, respectively?
My attempt was using the directional cosines, and I got that the square of the angle between the vector and the z-axis is negative, so it doesn't exist! However, is this a valid way to prove that this vector does not exist?

Is it true for $2\le <!--\; \le \; -->q\le <!--\; \le \; -->p$ that
$\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_q\le <!--\; \le \; -->n^{\frac{p-<!--\; -\; -->q}{pq}}\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_p$
where x is an n-dimensional vector. I only need the inequality for n=2, so that would suffice. I'm just curious if it's true for any n. Just from plugging it into a graphing calculator I believe it to be true, but how would I prove it?

Consider a matrix $X\in <!--\; \in \; -->{\mathbb{R}}_{n\times <!--\; \times \; -->m}$
One compact yet unclear notation to select a row or column from this matrix is:
$x\in <!--\; \in \; -->X$
How do you clearly select a row or column from a matrix?
I know $X=(x_{ij})$ is a standard notation to select elements. Though I haven't seen this used, $x_i\in <!--\; \in \; -->X_{ij}$ for rows and $x_j\in <!--\; \in \; -->X_{ij}$ for columns might make sense. This is motivated by a similar notation I have seen, namely $\underset{i}{\sum <!--\; \sum \; -->}{X}_{ij}$ for row sum or $\underset{j}{\sum <!--\; \sum \; -->}{X}_{ij}$ for column sum.

I have a linear map $R_{\mathrm{\alpha <!--\; \alpha \; -->}}=\left(\begin{array}{cc}cos\mathrm{\alpha <!--\; \alpha \; -->}& -<!--\; -\; -->sin\mathrm{\alpha <!--\; \alpha \; -->}\\ sin\mathrm{\alpha <!--\; \alpha \; -->}& cos\mathrm{\alpha <!--\; \alpha \; -->}\end{array}\right)$ and the question goes like, how does $R_{\mathrm{\alpha <!--\; \alpha \; -->}}$ transform a vector?
Can someone please help me?

I need to find the matrix of reflection through line $y=-<!--\; -\; -->\frac{2}{3}x$
I'm trying to visualise a vector satisfying this. The standard algorithm states that we need to find the angle this line makes with x axis and the transformation matrix can be seen as $R_{\mathrm{\alpha <!--\; \alpha \; -->}}{T}_0{R}_{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}}.$
I'm not sure how to proceed. I can't visualise the angle it makes with x axis. Is there a procedure to think about such reflections?

Show that $⟨<!--\; ⟨\; -->x+y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2^2-<!--\; -\; -->\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}_2^2$?
given $x,y\in <!--\; \in \; -->{\mathbb{R}}^n$ I am trying to prove this:
$⟨<!--\; ⟨\; -->x-<!--\; -\; -->y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}_2^2.$
What I tried is using the fact that:
$⟨<!--\; ⟨\; -->x-<!--\; -\; -->y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->y,y⟩<!--\; ⟩\; -->=\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2^2-<!--\; -\; -->2⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->+\Vert <!--\; \Vert \; -->y{\Vert <!--\; \Vert \; -->}_2^2.$
Then, since y=−(−y), I have
$⟨<!--\; ⟨\; -->x+y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->x-<!--\; -\; -->(-<!--\; -\; -->)y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->$
But I don't know how I proceed from this... Maybe my approach is wrong?

Consider the vectors $\mathbf{v}=\left(\begin{array}{c}1\\ 3\end{array}\right)$ and $\mathbf{w}=\left(\begin{array}{c}3\\ 2\end{array}\right)$
If the vectors v and w are linearly independent, answer with 0. If they aren't, find coefficients a and b, not both 0, such that $a\left(\begin{array}{c}1\\ 3\end{array}\right)+b\left(\begin{array}{c}3\\ 2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$ and answer with ${\displaystyle \frac{a}{b}}$
I know that vectors v and w are not linearly independent but im not sure how to get ${\displaystyle \frac{a}{b}}$

Can two n-dimensional vectors that are not scalar multiples of each other be added to make any point on an n-dimensional coordinate plan?
If we have $uu,vv\in <!--\; \in \; -->{\mathbb{R}}^n$ and $cuu\ne <!--\; \ne \; -->vv$ where $c\in <!--\; \in \; -->\mathbb{R}$, then is
${\mathrm{\forall <!--\; \forall \; -->}}_{a_1,\dots <!--\; \dots \; -->,a_n\in <!--\; \in \; -->\mathbb{R}}\left({\mathrm{\exists <!--\; \exists \; -->}}_{m,n\in <!--\; \in \; -->\mathbb{R}}(muu+nvv=\left[\begin{array}{c}{a}_1\\ ⋮<!--\; ⋮\; -->\\ a_n\end{array}\right])\right)\text{ true?}$
Note: Bolded letters represent vectors.

Are four non zero vectors are always linearly dependant
If three vectors are on the z-x plane and the fourth vector is not in this plane, the vectors would be linearly independent, no?