Vector Examples, Equations, and Practice Problems

Derive b from $\mathbf{a}=\mathbf{b}\times \mathbf{c}$
I have an equation:
$\mathbf{a}=\mathbf{b}\times \mathbf{c},$
where $\mathbf{a}$ $\mathbf{b}$ and $\mathbf{c}$ are 3-vectors.
How could I derive b from the equation and express it in terms of $\mathbf{a}$ and $\mathbf{c}$?

How can I generally solve equations of the form $\mathbf{A}\mathbf{w}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\times <!--\; \times \; -->\mathbf{w}$ for the matrix A, where w can be any vector? I recognize that you could just set w to a vector with simple values, such as $\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)$, but doing so still isn't helpful. Also, x, y, and z are entirely independent variables.

What is a mathematical explanation of the connection between: (1) projecting vector a onto vector b and multiplying the projected length of a with the length of vector b, and (2) the sum of the products of the equivalent components of the two vectors?
I realise there is a duality between a 2-dimensional vector and a 1x2 matrix, which can be used to explain the computation of the dot product. But I have not seen a satisfactory mathematical derivation, and was wondering whether there is another, simpler mathematical explanation.

How high would the object be thrown if three objects are to be juggled at 120 beats per minute( each object is thrown in turn one at a time and one per beat) with a horizontal displacement of approximately 0.30 m? use g = 9.8m/s

Let $\mathbb{F}$ be a field and suppose $a=(a_1,\dots <!--\; \dots \; -->,a_n)\in <!--\; \in \; -->{\mathbb{F}}^n$ and $b=(b_1,\dots <!--\; \dots \; -->,b_n)\in <!--\; \in \; -->{\mathbb{F}}^n$ are such that $\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k{b}_k=1$. Is it always possible to find $n\times <!--\; \times \; -->n$ matrices A and B such that the first row of A is a, the first column of B is b, and AB=I?
To me the answer is clearly yes but I am having some trouble proving this. I am fairly certain it is not needed but $\mathbb{F}$ is algebraically closed.
For those wondering what the context is for this problem, I am trying to determine whether two subalgebras are conjugate and I can construct an inner automorphism provided this fact always holds.

For a real vector I know we write $a=(a_1,a_2,a_3)\in <!--\; \in \; -->{\mathbb{R}}^3$, which means $a_1\in <!--\; \in \; -->\mathbb{R}$, $a_2\in <!--\; \in \; -->\mathbb{R}$ and $a_3\in <!--\; \in \; -->\mathbb{R}$
Suppose instead I have a vector $b=(b_1,b_2,b_3)$ where $b_1$, $b_2$ and $b_3$ only takes the values 0 or 1. What is the correct notation for this?

If a.i=4, then ,what is the value of (axj).(2j-3k) , where a is a vector

Given a vector-valued function defined by
${\displaystyle \mathbf{r}(t)=\left(\begin{array}{c}{t}^3+1\\ t^3+1\\ 2t+1\end{array}\right)}$
Let $\mathbb{T}$ denote the tangent to the curve at $A=(2,2,3)$
Then find the equation of the line L passing through the point u=(1,−1,2),parallel to the plane 2x+y+z=0 which intersects the tangent line $\mathbb{T}$
The equation of the line is in the form:
$\mathbb{L}=u+\stackrel{\to <!--\; \to \; -->}{v}s$
Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let $\stackrel{\to <!--\; \to \; -->}{n}=(2,1,1)$ denote the normal to the plane,then $\stackrel{\to <!--\; \to \; -->}{n}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{v}=(1,1,1)$, which implies:
$(v_2-<!--\; -\; -->v_3,2v_3-<!--\; -\; -->v_1,v_1-<!--\; -\; -->2v_2)=1$
So:
$\begin{array}{}\text{(1)}& v_2-<!--\; -\; -->v_3=1\end{array}$
$2v_3-<!--\; -\; -->v_1=1$
$\begin{array}{}\text{(2)}& v_1-<!--\; -\; -->2v_2=1\end{array}$
moreover $\stackrel{\to <!--\; \to \; -->}{n}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{v}=0$,which implies:
$\begin{array}{}\text{(3)}& 2v_1+v_2+v_3=0\end{array}$
Substituting (1) and (2) into (3) follows:
$v_1=2/3$
$v_2=-<!--\; -\; -->1/6$
$v_3=-<!--\; -\; -->7/6$
So the equation of the line is :
$\mathbb{L}=(1,-<!--\; -\; -->1,2)+(\frac{2}{3},-<!--\; -\; -->\frac{1}{6},-<!--\; -\; -->\frac{7}{6})s$
With the parametric equation :
$x=\frac{2}{3}s+1$
$y=-<!--\; -\; -->\frac{1}{6}s-<!--\; -\; -->1$
$z=-<!--\; -\; -->\frac{7}{6}s+2$
Since the line intersects the tangent line to the curve at a point with coordinate (2,2,3),we see that s=3/2,however substituting this to the y and z we don't get y=2 and z=3 respectively,so where was I wrong?

We want to find $O(x,y,z)$ by using two known points $P(2,3,5)$ and $Q(1,2,2)$ and angle of $POQ=60$ degrees with plane $E:1.15714286x+1.8547619y-<!--\; -\; -->z-<!--\; -\; -->2.86101191=0$ and length $OP=OQ$
Is there any way to find it?

Let T be the centroid of a triangle $\mathrm{△<!--\; △\; -->}ABC$
How can I prove that (using vectors):
$\stackrel{\to <!--\; \to \; -->}{TA}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{AB}=\stackrel{\to <!--\; \to \; -->}{TB}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{BC}=\stackrel{\to <!--\; \to \; -->}{TC}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{CA}$
($\times <!--\; \times \; -->$ is for vector-product).

Express the following plane in vector form:
${\mathcal{P}}_1\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^3$ with equation $4x-<!--\; -\; -->z=0$
The answer is $t(1,0,4)+s(0,1,0)$. I don't understand how they got (0,1,0) for the direction vector.

Question: Given $u=(2,4,-<!--\; -\; -->3),v=(3,-<!--\; -\; -->1,7),$ and i,j,k being the standard basis vector, find:
$|v-<!--\; -\; -->u+3k|$
My work:
$=|(3,-<!--\; -\; -->1,7)-<!--\; -\; -->(2,4,-<!--\; -\; -->3)+3k|$
$=|((3-<!--\; -\; -->2),(-<!--\; -\; -->1-<!--\; -\; -->4),7-<!--\; -\; -->(-<!--\; -\; -->3))+3k|$
$=|(1,-<!--\; -\; -->5,10)+3k|$
I know that the last line is not the full answer but I don't know what to do to continue to get the final result. I was thinking on making the coordinates (1,−5,10) into i−5j+10k and then adding the 10k with the 3k to get 13k but I am still confused on what to do with the absolute expression afterwards. If anyone can help me out, I'd appreciate it.

If I have $u=(u_1,u_2)$ is a vector of two components, then what is $\text{curl}u:=\mathrm{\nabla <!--\; \nabla \; -->}\times <!--\; \times \; -->u$? where $\mathrm{\nabla <!--\; \nabla \; -->}=({\mathrm{\partial <!--\; \partial \; -->}}_x,{\mathrm{\partial <!--\; \partial \; -->}}_y)$
I think it should give a scalar and not a vector.

Prove that $\stackrel{\to <!--\; \to \; -->}{u}$ and $\stackrel{\to <!--\; \to \; -->}{v}$ are orthogonal vectors
The vectors $\stackrel{\to <!--\; \to \; -->}{u}$ and $\stackrel{\to <!--\; \to \; -->}{v}$ are given in terms of the basis vectors $\stackrel{\to <!--\; \to \; -->}{a}$ , $\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{c}$ as follows:
$\stackrel{\to <!--\; \to \; -->}{u}=3\stackrel{\to <!--\; \to \; -->}{a}+3\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{c}$
$\stackrel{\to <!--\; \to \; -->}{v}=\stackrel{\to <!--\; \to \; -->}{a}+2\stackrel{\to <!--\; \to \; -->}{b}+3\stackrel{\to <!--\; \to \; -->}{c}$
I've tried $\stackrel{\to <!--\; \to \; -->}{u}.\stackrel{\to <!--\; \to \; -->}{v}$ to see if their dot product equals to 0, but it does not. Am I missing something?
It was given that $\stackrel{\to <!--\; \to \; -->}{a},\stackrel{\to <!--\; \to \; -->}{b},$ and $\stackrel{\to <!--\; \to \; -->}{c}$ form a basis in $R^3$.
It was also given that:
$|\stackrel{\to <!--\; \to \; -->}{a}|=1,|\stackrel{\to <!--\; \to \; -->}{b}|=2,|\stackrel{\to <!--\; \to \; -->}{c}|=3$

Question: Let $\stackrel{\to <!--\; \to \; -->}{u},\stackrel{\to <!--\; \to \; -->}{v}\in <!--\; \in \; -->{\mathbb{R}}^n$. $Span\{u⃗,v⃗\}=Span\{u⃗+v⃗,u⃗-v⃗\}$
I tried to approach this proof by finding a linear combination of $\stackrel{\to <!--\; \to \; -->}{u}$ and $\stackrel{\to <!--\; \to \; -->}{v}$ but I am confused as to how to approach the linear combination of the right hand side of the equation. Please help!

I'm trying to find the magnitude of a vector $\stackrel{\dot{}<!--\; \dot{}\; -->}{\stackrel{\to <!--\; \to \; -->}{r}}(t)$ by two different methods, but get different results.
Let $\stackrel{\to <!--\; \to \; -->}{r}(t)=x(t)\stackrel{^<!--\; ^\; -->}{x}+y(t)\stackrel{^<!--\; ^\; -->}{y}$. The magnitude of this vector is:
$||\stackrel{\to <!--\; \to \; -->}{r}(t)||=\sqrt{x(t{)}^2+y(t{)}^2}$
The time derivative of $\stackrel{\to <!--\; \to \; -->}{r}(t)$ must then be
$\stackrel{\dot{}<!--\; \dot{}\; -->}{\stackrel{\to <!--\; \to \; -->}{r}}(t)=\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)\stackrel{^<!--\; ^\; -->}{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}+\stackrel{\dot{}<!--\; \dot{}\; -->}{y}(t)\stackrel{^<!--\; ^\; -->}{\stackrel{\dot{}<!--\; \dot{}\; -->}{y}}$
Now, using the same formula for the magnitude as before yields:
$||\stackrel{\dot{}<!--\; \dot{}\; -->}{\stackrel{\to <!--\; \to \; -->}{r}}(t)||=\sqrt{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t{)}^2+\stackrel{\dot{}<!--\; \dot{}\; -->}{y}(t{)}^2}$
But when I try to find the magnitude of $\stackrel{\dot{}<!--\; \dot{}\; -->}{\stackrel{\to <!--\; \to \; -->}{r}}(t)$ by differentiating $||\stackrel{\to <!--\; \to \; -->}{r}(t)||$ with respect to time, the chain rule gives me:
$\begin{array}{rl}\frac{d}{dt}||\stackrel{\to <!--\; \to \; -->}{r}||& =\frac{\mathrm{\partial <!--\; \partial \; -->}||\stackrel{\to <!--\; \to \; -->}{r}||}{\mathrm{\partial <!--\; \partial \; -->}x}\frac{dx}{dt}+\frac{\mathrm{\partial <!--\; \partial \; -->}||\stackrel{\to <!--\; \to \; -->}{r}||}{\mathrm{\partial <!--\; \partial \; -->}y}\frac{dy}{dt}\\ & =\frac{2x}{2\sqrt{x^2+y^2}}\stackrel{\dot{}<!--\; \dot{}\; -->}{x}+\frac{2y}{2\sqrt{x^2+y^2}}\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\\ & =\stackrel{\dot{}<!--\; \dot{}\; -->}{x}\frac{x}{||r||}+\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\frac{y}{||r||}\ne <!--\; \ne \; -->\sqrt{{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^2+{\stackrel{\dot{}<!--\; \dot{}\; -->}{y}}^2}\end{array}$
Where it is implicit that all the functions are evaluated in t. Have I made an algebraic mistake, or is it not possible to find $||\stackrel{\dot{}<!--\; \dot{}\; -->}{\stackrel{\to <!--\; \to \; -->}{r}}(t)||$ by differentiation? - and if so, why?

Determine whether the vector field F is conservative or not. If it is, find its potential function (i.e., find a real-valued function V such that F = -grad V).
$F(x,y,z)=xi+y^{2}j+zk$

Does it mathematically make sense for a number to be added to a vector (or matrix for that matter)? When I put $4+[1,2,3]$ into Wolfram Alpha it gives me $[5,6,7]$, i.e. the number is added to each item, but this doesn't seem correct to me as I would have thought that the number would just be added to the 1st item of the vector - or is neither correct?

Calculate the norm of vectors (1,0) and (0,1). We are on ${\mathbb{R}}^2$ where $⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->=2x_1{y}_1+x_2{y}_2$. I know that I need to calculate the norm by $||x||=\sqrt{⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->}$, but I'm having difficulty using that.
Looking at the first one, some of my classmates are computing it as $\sqrt{2\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->1+0\cdot <!--\; \cdot \; -->0}$, but I don't understand where that comes from.
Personally, I was thinking it would be $\sqrt{2\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->1+1\cdot <!--\; \cdot \; -->1}$, but honestly I'm not super confident there.

Let V be a $\mathbb{K}$-Vectorspace, and let $B_1,B_2,...$ be subsets of V such that $B_1\supseteq <!--\; \supseteq \; -->B_2\supseteq <!--\; \supseteq \; -->...$ and for all $i\in <!--\; \in \; -->\mathbb{N}\setminus <!--\; \setminus \; -->\{0\}$ is $⟨<!--\; ⟨\; -->B_i⟩<!--\; ⟩\; -->=V$. Prove or disprove that $⟨<!--\; ⟨\; -->{\cap <!--\; \cap \; -->}_{i\in <!--\; \in \; -->\mathbb{N}}{B}_i⟩<!--\; ⟩\; -->=V$. The hint from our professor was that this is not true. Let's choose $B_k$ such that it is a subset of all other $B_i$ and also the smallest set. So every vector of $B_k$ is also in every other $B_i$. Therefore, the intersection of all $B_i$ would be $B_k$ and we know that $⟨<!--\; ⟨\; -->B_k⟩<!--\; ⟩\; -->=V$ which would make the statement true. Where am I wrong? Any solution or hint would be highly appreciated.