Vector Examples, Equations, and Practice Problems

Two lines are given by the equation x = 10 - 8t, y=1+8t, z=15 +2t and x=6-8t, y=2+2t, z=3+6t What is the shortest distance between these two lines?

I have the points O,A,B and C.
Relative to O, the position vectors of A, B and C are (1,4,2), (3,3,3), (2,−1,1)
Want to show that the lines OB and AC bisect each other.
Is it sufficient to show that $\frac{1}{2}\stackrel{\to <!--\; \to \; -->}{OB}=\stackrel{\to <!--\; \to \; -->}{OA}+\frac{1}{2}\stackrel{\to <!--\; \to \; -->}{AC}$?
Are there other ways using vectors?

I have got this line
$\mathrm{g}:\stackrel{\to <!--\; \to \; -->}{\mathrm{x}}=\left(\begin{array}{c}-<!--\; -\; -->1\\ 2\\ 0\end{array}\right)+\mathrm{s}\cdot <!--\; \cdot \; -->\left(\begin{array}{c}-<!--\; -\; -->1\\ 4\\ -<!--\; -\; -->1\end{array}\right)$
And this plane:
$E_t:tx+y+tz=0$
Now for which t is the line element of the plane?
My calculations: When I put the equation of the line in the equation of the plane i get: And this plane:
$t(-<!--\; -\; -->1-<!--\; -\; -->s)+(2+4s)+t(-<!--\; -\; -->2)=0$
This is the same as:
$2-<!--\; -\; -->t+s(4-<!--\; -\; -->2t)=0$
$=>s=\frac{t-<!--\; -\; -->2}{4-<!--\; -\; -->2t}=-<!--\; -\; -->\frac{1}{2}$
And now I do not know how to go on.
The solution is t=2
Can someone explain the last step to the solution please?

How can we prove that "We know that for any vector x, $x^{T}x=||x|{|}^2$. Thus , ..... "
The way I am thinking that while $x^T$ is the transpose of x, then we cross product with itself using $x^T$, which results in a symmetric matrix, R. So R is a square matrix. How can we jump from a square matrix to $||x|{|}^2,$, where ||x|| supposed to mean the normal of x, then we square it with 2?

Why can $v_x,v_y$ in $\mathbf{v}=(z,z,z)$ contain the z variable?

Let $\{v_1,v_2,v_3,v_4\}$ be a vector basis of ${\mathbb{R}}^4$ and A a constant matrix of ${\mathbb{R}}^{4\times <!--\; \times \; -->4}$ so that:
$A{v}_1=-<!--\; -\; -->2v_1,A{v}_2=-<!--\; -\; -->v_1,A{v}_3=3v_4,A{v}_4=-<!--\; -\; -->3v_3$
Can I find the eigenvalues of the matrix A? I know that ${\mathrm{\lambda <!--\; \lambda \; -->}}_1=-<!--\; -\; -->2$ is a trivial eigenvalue but I don't know how to calculate the others.

Let u and v be two vectors in ${\mathbb{R}}^2$
The Cauchy-Schwarz inequality states that $|u·v|\le |u||v|$
We are able to transform the above inequality so that it also shows us that
$|u+v|\le |u|+|v|$
But I cannot find a way to show that $\mathbf{|}\mathbf{u}\mathbf{-<!--\; -\; -->}\mathbf{v}\mathbf{|}\mathbf{\le }\mathbf{|}\mathbf{u}\mathbf{|}\mathbf{+}\mathbf{|}\mathbf{v}\mathbf{|}$
even though I now this has to be true. Any ideas?

I need to find $\frac{dy}{dx}$ for the following
y = $||A^{T}x-<!--\; -\; -->b|{|}_2^2$ where $A\in <!--\; \in \; -->R^{3x3},b\in <!--\; \in \; -->R^{3x1},x\in <!--\; \in \; -->R^{3x1},y\in <!--\; \in \; -->R,$ and $||.|{|}_2$ is the euclidean norm so for example $||z|{|}_2^2=z^{T}z$ for $z\in <!--\; \in \; -->R^{3x1}$. I'm familiar with the chain rule but I've never really used it in this way. Also, I'm not sure what $R^{3x3}$ represents and how I can use it with the chain rule.

Show that $a=2i+2j+3k,b=3i+j-<!--\; -\; -->k,c=i-<!--\; -\; -->j-<!--\; -\; -->4k$ forms the sides of a triangle.
My attempt: $|a|=\sqrt{17},|b|=\sqrt{11},|c|=\sqrt{18}.$ Since $|c|<|a|+|b|$ using triangle inequality, we can say a,b,c form sides of a triangle.
I am not sure if my attempt is correct.

Help in understanding properties of big O notation with norms of vectors
In Fast Exact Multiplication by the Hessian equation 1,
$O({\Vert \mathrm{\Delta <!--\; \Delta \; -->}w\Vert }^2)$ gets taken from RHS to LHS and $\mathrm{\Delta <!--\; \Delta \; -->}w$ is substituted as rv where r is small scalar and v is a vector. I understand that $O({\Vert rv\Vert }^2)=O(r^2{\Vert v\Vert }^2)$. But what I don't get is how did the sign of O not become negative when it was taken to LHS. And why did ${\Vert v\Vert }^2$ disappear form O. Is it because as r is tiny it will govern the O term and ${\Vert v\Vert }^2$ won't matter?

Proof that a subset of ${\mathbb{R}}^2$ is not a vector space
Here is a subset of ${\mathbb{R}}^2$:
$A=[(x_1,x_2)\in <!--\; \in \; -->{\mathbb{R}}^2:x_1-<!--\; -\; -->x_2^2=0]$
I am trying to prove that it is not a vector subspace.
It is not empty, the (0,0) vector works
It is closed under scalar multiplication (sign of $x_1$ and $x_2$ are changed but it works)
I was a bit struggling to prove that it does not work for all $x_1$ and $x_2$ (Is it possible ?)
I have chosen to take an u=(1,−1) and v=(16,4). Both of them are in ${\mathbb{R}}^2$. When I do u+v, I got a vector that is (17,3). This vector is not in A since $x_1-<!--\; -\; -->x_2^2\ne <!--\; \ne \; -->0$
My question : Is there something wrong in my reasoning / solution ?

Let $\stackrel{\to <!--\; \to \; -->}{a}=i+2j+3k$ and $\stackrel{\to <!--\; \to \; -->}{b}=2i+5k$. For which value of t when $-2\le t\le 2$ holds the length of the vector $\stackrel{\to <!--\; \to \; -->}{c_t}=t\stackrel{\to <!--\; \to \; -->}{a}+(1-t)\stackrel{\to <!--\; \to \; -->}{b}$ is as small as possible?
How should one approach this? The minimum would be when $\stackrel{\to <!--\; \to \; -->}{c_t}$ is perpendicular to some vector $\stackrel{\to <!--\; \to \; -->}{d}=\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}$ or is there something else Im not seeing?

How to find all points for which the tangent line of a parametric equation (x, y, z) passes through a point
For the function
$\stackrel{\to <!--\; \to \; -->}{x}(t)=\left(\begin{array}{c}2t+3\\ 2-<!--\; -\; -->t\\ t^3-<!--\; -\; -->2t^2+t\end{array}\right)t\ge 0$
Find all points $\stackrel{\to <!--\; \to \; -->}{x}(t_0)$ for which the tangent line passes through the point (1, 3, 0).
I understand how I would go about solving the problem if I had a normal polynomial function or only x and y values. However, with x, y, and z values I am not sure how to approach the question.
I know that the derivative of the function is
${\stackrel{\to <!--\; \to \; -->}{x}}^{\prime }(t)=\left(\begin{array}{c}2\\ -<!--\; -\; -->1\\ 3t^2-<!--\; -\; -->4t+1\end{array}\right)$
Would I then have to find the equation of the tangent line and use that with the given point to find the other points?

I need to find the angle between two unit vectors $\stackrel{\to <!--\; \to \; -->}{m}$ and $\stackrel{\to <!--\; \to \; -->}{n}$ if the vectors $\stackrel{\to <!--\; \to \; -->}{p}=\stackrel{\to <!--\; \to \; -->}{m}+2\stackrel{\to <!--\; \to \; -->}{n}$ and $\stackrel{\to <!--\; \to \; -->}{q}=5\stackrel{\to <!--\; \to \; -->}{m}-<!--\; -\; -->4\stackrel{\to <!--\; \to \; -->}{n}$ are perpendicular to each other.

Help me with my assignment which has a notation like this: $P_1=(x,y,z)\in <!--\; \in \; -->{\mathbb{R}}^3:|x|\le <!--\; \le \; -->1,|y|\le <!--\; \le \; -->1,|z|\le <!--\; \le \; -->1$. At first I thought $|x|=\sqrt{{x_1}^2+{x_2}^2+{x_3}^2}$, which is the magnitude of 3D vector x. I read in many sources saying there's no absolute value for a vector, there's only $||x||=\sqrt{{x_1}^2+{x_2}^2+{x_3}^2}$. So I want to ask if what |x| actually is and how its formula looks like?

I was trying to give a geometrical description of the the motion of Q, and the motion of Q can be described by the equation: $Q=cos(t+\frac{1}{4}\mathrm{\pi <!--\; \pi \; -->})[\frac{3}{2}\stackrel{\to <!--\; \to \; -->}{i}+\frac{3\sqrt{3}}{2}\stackrel{\to <!--\; \to \; -->}{k}]+3sin(t+\frac{1}{4}\mathrm{\pi <!--\; \pi \; -->})\stackrel{\to <!--\; \to \; -->}{j}$
By calculating |Q|, we know that the distance from the origin is constant; hence, the particle would travel at a circular path.After proving that, I took a look at the mark scheme for this question, and the mark scheme indicated that:”it is evident that $\sqrt{3}x-<!--\; -\; -->z=0$, and so this defines the plane in which the motion of Q takes place”. I can’t understand what $\sqrt{3}x-<!--\; -\; -->z=0$ tells us, and why does it relate to the plane of motion, from my point of view, we can see that the equation of motion has variables in x-y-z, so it is obviously moves in x-y-z.

Let's say I'm given a vector as follows:
$x=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$
And I'd like to produce the following matrix:
$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$
What series of operations can I induce to produce this matrix?

Let $\theta$ be the angle between the vectors $A=(1,1,...,1)$ and $B=(1,2,...,n)$ in ${\mathbb{R}}^n$. Find the limiting value of $\mathrm{\theta <!--\; \theta \; -->}$ as $b\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$
For this question, I want to apply the equation for the angle between two vectors: $\theta =arccos\frac{A·B}{\Vert <!--\; \Vert \; -->A\Vert <!--\; \Vert \; -->\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$. For A·B, I use $A·B=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k{b}_k$. And for the norm of A and B, I simply use the definition to get $\Vert <!--\; \Vert \; -->A\Vert <!--\; \Vert \; -->=\sqrt{n}$ and $\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->=\sqrt{\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{k}^2}$. But I'm stuck here. I don't know how to calculate the limit of $\mathrm{\theta <!--\; \theta \; -->}$ as $b\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ by these equations.

Every vector v can be expressed uniquely in the form $\mathbf{a}+\mathbf{b},$, where $\mathbf{a}$ is a scalar multiple of $\left(\begin{array}{c}2\\ -<!--\; -\; -->1\end{array}\right),$, and b is a scalar multiple of $\left(\begin{array}{c}3\\ 1\end{array}\right).$. Find the matrix $\mathbf{P}$ such that $\mathbf{P}\mathbf{v}=\mathbf{a}$ for all vectors $\mathbf{v}.$
I'd like help deriving $\mathbf{P}$, but I don't know how to do it. Any help would be much appreciated!

I ran into some trouble understanding how my professor went from this step in his solution:
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}{\left|\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{imx}-<!--\; -\; -->\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{inx}\right|}^2\mathrm{d}x$
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}{\left|\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{imx}-<!--\; -\; -->\frac{1}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{inx}\right|}^2\mathrm{d}x$
to this step:
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}(1-<!--\; -\; -->e^{i(m-<!--\; -\; -->n)x}-<!--\; -\; -->e^{i(n-<!--\; -\; -->m)x}+1)\mathrm{d}x$
I cannot see how this was done, please help!