Vector Examples, Equations, and Practice Problems

Find $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->\left(\frac{\mathit{r}}{r^2}\right)$ where $\mathit{r}(x,y,z)=x\mathit{i}+y\mathit{j}+z\mathit{k}$, $r=\sqrt{x^2+y^2+z^2}$
In this case, why can we not bring the scalar function $r^{-<!--\; -\; -->2}$ outside the dot product?
When I leave the function in place, I calculate a result of $\frac{1}{r^2}$. When I pull it out, I find $\frac{3}{r^2}$

Find the vector set of all orthogonal to vectors $\stackrel{\to <!--\; \to \; -->}{v}=(1,2,3,4)$, $\stackrel{\to <!--\; \to \; -->}{w}=(2,-<!--\; -\; -->2,6,-<!--\; -\; -->4)$ in ${\mathbb{R}}^4$. I think orthogonal is means the vectors dot product to be 0. I maybe can find one of the vectors, but how can I find all of them?

i know i can write the geodesic equation for a massive particle as:
${\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\nabla <!--\; \nabla \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}}{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^{\mathrm{\mu <!--\; \mu \; -->}}=0$
and then we can express this using the 4 momentum, $p^{\mathrm{\mu <!--\; \mu \; -->}}=m{u}^{\mathrm{\mu <!--\; \mu \; -->}}=m{\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}^{\mathrm{\mu <!--\; \mu \; -->}}$
$p^{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\nabla <!--\; \nabla \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}}{p}^{\mathrm{\mu <!--\; \mu \; -->}}=0$
I want to show that this can be written as
$m\frac{d{p}_{\mathrm{\mu <!--\; \mu \; -->}}}{d\mathrm{\tau <!--\; \tau \; -->}}=\frac{1}{2}{p}^{\mathrm{\sigma <!--\; \sigma \; -->}}{p}^{\mathrm{\rho <!--\; \rho \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}}{g}_{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\sigma <!--\; \sigma \; -->}}$
i expanded the covariant derivative out and lowered some of the indices using the metric to obtain that
$g^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\rho <!--\; \rho \; -->}}{p}^{\mathrm{\nu <!--\; \nu \; -->}}[{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}}{p}_{\mathrm{\rho <!--\; \rho \; -->}}-<!--\; -\; -->{\mathrm{\Gamma <!--\; \Gamma \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}\mathrm{\rho <!--\; \rho \; -->}}^{\mathrm{\alpha <!--\; \alpha \; -->}}{p}_{\mathrm{\alpha <!--\; \alpha \; -->}}]=0$
the first term in that expression can be written as
$g^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\rho <!--\; \rho \; -->}}m\frac{\mathrm{\partial <!--\; \partial \; -->}{p}_{\mathrm{\rho <!--\; \rho \; -->}}}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\tau <!--\; \tau \; -->}}$
and i know i can get a 1/2 out of the christofell symbol but everything i try ends in a mess of metrics derivatives and indices.

The matrix A has size $3\times <!--\; \times \; -->3$ and we know that for any column vector $v\in <!--\; \in \; -->{\mathbb{R}}^3$ the vectors $Av$ and v are orthogonal. Prove that $A^T+A=0$, where $A^T$ is the transposed matrix $A$
So if
$A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)$
and
$v=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$
orthogonality of Av and v brought me to the equation
$x\cdot <!--\; \cdot \; -->y\cdot <!--\; \cdot \; -->(a_{12}+a_{21})+x\cdot <!--\; \cdot \; -->z\cdot <!--\; \cdot \; -->(a_{13}+a_{31})+y\cdot <!--\; \cdot \; -->z\cdot <!--\; \cdot \; -->(a_{23}+a_{32})+a_{11}\cdot <!--\; \cdot \; -->x^2+a_{22}\cdot <!--\; \cdot \; -->y^{22}+a_{33}\cdot <!--\; \cdot \; -->z^2=0$
and the matrix $A^T+A$ equals
$A^T+A=\left(\begin{array}{ccc}2a_{11}& a_{12}+a_{21}& a_{13}+a_{31}\\ a_{12}+a_{21}& 2a_{22}& a_{23}+a_{32}\\ a_{13}+a_{31}& a_{23}+a_{32}& 2a_{33}\end{array}\right)$
However I don't see how then prove that $A^T+A=0$

Let $\stackrel{\to <!--\; \to \; -->}{a},\stackrel{\to <!--\; \to \; -->}{b}\in <!--\; \in \; -->{\mathbb{R}}^3$ be given,
Consider the curve $\stackrel{\to <!--\; \to \; -->}{r}(t)=\stackrel{\to <!--\; \to \; -->}{O{P}_0}+t\stackrel{\to <!--\; \to \; -->}{a}+t^2\stackrel{\to <!--\; \to \; -->}{b}$ , $t\in <!--\; \in \; -->\mathbb{R}$
Find the curvature of the curve at $\stackrel{\to <!--\; \to \; -->}{r}(0)$
I have just started learning about arc-lengths and paremetrizations, and I have no clue how to approach this question. I've tried splitting each vector in the function to its components (e.g. $\stackrel{\to <!--\; \to \; -->}{a}\to <!--\; \to \; -->(a_1,a_2,a_3))$, but I can't progress any further. I'm wondering how to incorporate the curvature formula for this question;
$\frac{{\stackrel{\to <!--\; \to \; -->}{r}}^{\prime }(t)\times <!--\; \times \; -->{\stackrel{\to <!--\; \to \; -->}{r}}^{″}(t)}{{\left|\stackrel{\to <!--\; \to \; -->}{r(t)}\right|}^3}$
Any help would be greatly appreciated.

If a,b,c are three non coplanar vectors, then prove that the vector equation
$r=(1-<!--\; -\; -->p-<!--\; -\; -->q)\stackrel{\to <!--\; \to \; -->}{a}+p\stackrel{\to <!--\; \to \; -->}{b}+q\stackrel{\to <!--\; \to \; -->}{c}$ represents a plane
Let $r=x\stackrel{\to <!--\; \to \; -->}{a}+y\stackrel{\to <!--\; \to \; -->}{b}+z\stackrel{\to <!--\; \to \; -->}{c}$
Comparing with the given equation, we obtain
x+y+z=1
which is a plane
What I don’t understand is how does this say r is a plane, since r is actually $x\stackrel{\to <!--\; \to \; -->}{a}+y\stackrel{\to <!--\; \to \; -->}{b}+z\stackrel{\to <!--\; \to \; -->}{c}$

How do I find the dot product of $\stackrel{\to <!--\; \to \; -->}{u}+\stackrel{\to <!--\; \to \; -->}{v}$ and $\stackrel{\to <!--\; \to \; -->}{u}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{v}$ with the given information?
I already know that $|\stackrel{\to <!--\; \to \; -->}{u}|=2,|\stackrel{\to <!--\; \to \; -->}{v}|=3,|$ and $⟨<!--\; ⟨\; -->\stackrel{\to <!--\; \to \; -->}{u},\stackrel{\to <!--\; \to \; -->}{v}⟩<!--\; ⟩\; -->=1$. I am unsure as to how to proceed in order to find $⟨<!--\; ⟨\; -->\stackrel{\to <!--\; \to \; -->}{u}+\stackrel{\to <!--\; \to \; -->}{v},\stackrel{\to <!--\; \to \; -->}{u}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{v}⟩<!--\; ⟩\; -->$

Why is the vector $(I-<!--\; -\; -->A{)}^{-<!--\; -\; -->1}x$ positive?
I was reading a text where they make the conclusion that the product $(I-<!--\; -\; -->A{)}^{-<!--\; -\; -->1}x$ is a vector with positive entries.
Notes:
1)I is the identity $n\times <!--\; \times \; -->n$ matrix.
2)A is a $n\times <!--\; \times \; -->n$ matrix with zeros on the diagonal and non-negative entries such that $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}<1$, thus $I-<!--\; -\; -->A$ is invertible.
3)x is a vector with positive entries.

Given a point $P=[x,y,z]$ and a vector $v=[v_x,v_y,v_z]$
I want to move the point along the vector by a fixed amount d

For scalars a,b we know that $(a-<!--\; -\; -->b{)}^2\ge <!--\; \ge \; -->0⟹<!--\; ⟹\; -->2a^2+2b^2\ge <!--\; \ge \; -->(a+b{)}^2$
Is the analogous inequality for vectors also always true?
i.e., does $||\underset{i}{\sum <!--\; \sum \; -->}{a}_i|{|}^2\le <!--\; \le \; -->2\underset{i}{\sum <!--\; \sum \; -->}||a_i|{|}^2$ ?
Edit:
We can show similarly for 2 vectors that $||a+b|{|}^2\le <!--\; \le \; -->2||a|{|}^2+2||b|{|}^2$
Now, for 3 vectors, we have $||a+b+c|{|}^2\le <!--\; \le \; -->2||a+b|{|}^2+2||c|{|}^2\le <!--\; \le \; -->4||a|{|}^2+4||b|{|}^2+2||c|{|}^2$
Is there a general form? like $||\underset{i}{\sum <!--\; \sum \; -->}{a}_i|{|}^2\le <!--\; \le \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_i||a_i|{|}^2$

Given $|\stackrel{\to <!--\; \to \; -->}{x}|=2,|\stackrel{\to <!--\; \to \; -->}{y}|=3$ and the angle between them is 120°, determine the unit vector in the opposite direction of $|\stackrel{\to <!--\; \to \; -->}{x}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{y}|$

I am trying to show that if $span\{2x+y,y+z\}=span\{x,y,z\}$ then $z\in <!--\; \in \; -->span\{2x+y,y+z\}$ where x and y are non-zero vectors.
Let $a,b,c,d,e\in <!--\; \in \; -->\mathbb{F}$ where $\mathbb{F}$ is field of our vector space.
We can write $a(2x+y)+b(y+z)=cx+dy+ez$ by the equality of sets. However I could not write z as a linear combination of $2x+y$ and $y+z$
I am so sorry if there is a mistake.
Thanks in advance for any help

Maximum value of $|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{b}{|}^2+|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{c}{|}^2+|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{a}{|}^2$
I have a question the goes, if $|\stackrel{\to <!--\; \to \; -->}{a}|=|\stackrel{\to <!--\; \to \; -->}{b}|=1$ and $|\stackrel{\to <!--\; \to \; -->}{c}|=2$, then what is the maximum value of :
$|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{b}{|}^2+|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{c}{|}^2+|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{a}{|}^2$?
Here's what I tried:
$|\stackrel{\to <!--\; \to \; -->}{a}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{b}{|}^2+|\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{c}{|}^2+|\stackrel{\to <!--\; \to \; -->}{c}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{a}{|}^2=30-<!--\; -\; -->4\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1-<!--\; -\; -->8\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_2-<!--\; -\; -->8\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_3$
where ${\mathrm{\theta <!--\; \theta \; -->}}_1=$ angle between $\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{b}$ , ${\mathrm{\theta <!--\; \theta \; -->}}_2=$ angle between $\stackrel{\to <!--\; \to \; -->}{b}$ and $\stackrel{\to <!--\; \to \; -->}{c}$, ${\mathrm{\theta <!--\; \theta \; -->}}_3=$ angle between $\stackrel{\to <!--\; \to \; -->}{c}$ and $\stackrel{\to <!--\; \to \; -->}{a}$
and for maximum value, I put $\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1=\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_2=\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_3=-<!--\; -\; -->1$, and got the maximum value of 50 but the answer says 42. Where did I go wrong and what would be the correct approach? Was I wrong in putting $\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_1=\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_2=\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}_3=-<!--\; -\; -->1$? Why?

I need to prove that for 0${\Vert x\Vert }_p$ = ($\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|x_i{|}^p$)${}^{1/p}$ for $n\ge <!--\; \ge \; -->2$
no longer satisfies the triangle inequality?
I know that in one dimension, the norm is just an absolute value, and the triangle inequality holds. I could take two unit vectors.
For example, $\stackrel{\to <!--\; \to \; -->}{a}$ = (0,1,0) and $\stackrel{\to <!--\; \to \; -->}{b}$ = (2/3.2/3.1/3)
But I don't really know how I can put the unit vectors in the sum above to show that
$\Vert a+b\Vert$ $\le <!--\; \le \; -->$ $\Vert a\Vert$ + $\Vert b\Vert$ does not hold.

I have a question about the solution to this problem ''find the flux of $x\stackrel{^<!--\; ^\; -->}{i}+y\stackrel{^<!--\; ^\; -->}{j}+z\stackrel{^<!--\; ^\; -->}{k}$ through the sphere of radius a and center at the origin. Take n pointing outward.''
The answer in the book was, we have $n=\frac{(x\stackrel{^<!--\; ^\; -->}{i}+y\stackrel{^<!--\; ^\; -->}{j}+z\stackrel{^<!--\; ^\; -->}{k})}{a}$; therefore $F.n=a$ and then they integrate it, but what I don't get is how $F.n=a$ isn't the vector n the same vector as $F$ but scaled by $1/a$ so the dot product must be $\frac{(x^2\stackrel{^<!--\; ^\; -->}{i}+y^2\stackrel{^<!--\; ^\; -->}{j}+z^2\stackrel{^<!--\; ^\; -->}{k})}{a}$

this is my question, any help would be appreciated.
Consider the vector $D=3\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ ar (unit vector).
Evaluate the integral
${\iint <!--\; \iint \; -->}_{S}DdS,$
where S is the surface of a sphere with radius r=5 centered at the origin.

I really couldn't find anything related to this simple identity I came up with so:
$\stackrel{\to <!--\; \to \; -->}{r}=(r_x,r_y)=(r_x,\mathrm{\angle <!--\; \angle \; -->}0)+(r_y,\mathrm{\angle <!--\; \angle \; -->}\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$
My thinking process was that $r_y$ is practically the modulus of a vector in the Y axis (or with $\mathrm{\theta <!--\; \theta \; -->}=90°=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$) and $r_x$ is the modulus of a vector in the X axis (or with $\mathrm{\theta <!--\; \theta \; -->}=0°=0$).
Is this true?

Consider the surface $$z=(\frac{x}{2}{)}^2-<!--\; -\; -->(\frac{y}{3}{)}^2$$. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

How would I go about finding a vector equation of a hyperplane
$3x_1+x_2-<!--\; -\; -->2x_3+4x_4=5$
I know I need to find 3 vectors and another vector that shifts from the origin in the form
$r\stackrel{\to <!--\; \to \; -->}{a}+s\stackrel{\to <!--\; \to \; -->}{b}+t\stackrel{\to <!--\; \to \; -->}{c}+\stackrel{\to <!--\; \to \; -->}{d}$

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Under what conditions, if any, does a vector $(u,v,w)$ lie under the image of T?
I'm being asked to consider $T:{\mathbb{R}}^3\to <!--\; \to \; -->{\mathbb{R}}^3$ given by the formula:
$(u,v,w)=T(x,y,z)=(x-<!--\; -\; -->y,y-<!--\; -\; -->z,z-<!--\; -\; -->x)$
Then, I'm asked under what conditions, if any, does a vector (u,v,w) lie in the image of T? lie under the image of T?
If my understanding is correct, the term "image" refers to the range of a transformation, meaning any element mapped to by T within the codomain is the image. lie under the image of T?
It looks to me that any value for u, v, or w will be acceptable under T, but I don't know how to validate that. I think I need some sort of equation that will prove that any input will produce a valid output. Is this a valid line of thinking, or am I on the wrong track? lie under the image of T?
My question is: How do I prove that any value is acceptable for a valid output?