Vector Examples, Equations, and Practice Problems

Is ${\mathbb{Q}}^n$ a vector space over $\mathbb{Z}$ or over $\mathbb{Q}$?
${\mathbb{Q}}^n$ is clearly not a vector space over $\mathbb{R}$, because scalar multiplication of some $q\in <!--\; \in \; -->\mathbb{Q}$ by $\mathrm{\pi <!--\; \pi \; -->}$ renders $\mathrm{\pi <!--\; \pi \; -->}q$, which is irrational.

Show that an equation of a plane passing through three noncolinear points $p_1=(x_1,y_1,z_1)$, $p_2=(x_2,y_2,z_2)$, $p_3=(x_3,y_3,z_3)$ is given by $(p-p_1)\times <!--\; \times \; -->(p-p_2)\cdot <!--\; \cdot \; -->(p-p_3)=0$, where p=(x,y,z) is an arbitrary point of the plane and $p-p_1$, for instance, means the vector $(x-x_1,y-y_1,z-z_1)$
I have the following reasoning:
"Explicitly, the plane $\mathcal{P}$ through the point $P_0=(x_0,y_0,z_0)$ is uniquely determined (up to a scalar multiple) by a normal vector n=⟨a,b,c⟩ according to the following: a point P lies on $\mathcal{P}$ if and only if n and $\stackrel{\to <!--\; \to \; -->}{P_{0}P}$ are orthogonal if and only if $n\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{P_{0}P}=0$ if and only if
$a(x-<!--\; -\; -->x_0)+b(y-<!--\; -\; -->y_0)+c(z-<!--\; -\; -->z_0)=0.$
By setting $d=a{x}_0+b{y}_0+c{z}_0,$ we have $ax+by+cz=d.$
But I have not been able to finish the problem because I must get the equation of the plane of the form $(p-p_1)\times <!--\; \times \; -->(p-p_2)\cdot <!--\; \cdot \; -->(p-p_3)=0$. I need help to do this.

How do I find the vector length for high dimensions?.We can find vector length for 3d with the formula $\sqrt{v_1^2+v_2^2+v_3^2}$ Likewise how to find the vector magnitude for high dimensions?

Three vectors, $V_1,V_2,V_3$ are in the ${\mathbb{R}}^2$ plane where $V_1+V_2+V_3=\stackrel{\to <!--\; \to \; -->}{0}$ and the magnitudes of these vectors are the same. Show that the angle between any two of these vectors is 120 degrees.
My try:
$V_1+V_2+V_3=\stackrel{\to <!--\; \to \; -->}{0}$, so $V_1+V_2=-<!--\; -\; -->V_3$. This would mean that $(V_1+V_2{)}^2=V_3^2$ which is $|V_1|+|V_2|+2V_1{V}_2=|V_3|$. But since the magnitude is the same, we get $2|V_1|+2V_1{V}_2=|V_1|$ which is $2V_1{V}_2=-<!--\; -\; -->|V_1|$
The formula for the angle is $\cos <!--\; \; -->(a)=\frac{V_1\ast <!--\; \ast \; -->V_2}{|V_1|\ast <!--\; \ast \; -->|V_2|}=-<!--\; -\; -->0.5\ast <!--\; \ast \; -->\frac{|V_1|}{|V_1{|}^2}=\frac{-<!--\; -\; -->0.5}{|V1|}$
I know that the solution would be a triangle and that $\cos <!--\; \; -->(a)=-<!--\; -\; -->0.5$
But that would mean that my angle is dependent on $V_1$ which isn't the case. So what am I doing wrong?

Prove $u\cdot <!--\; \cdot \; -->(u\times <!--\; \times \; -->(\mathrm{\nabla <!--\; \nabla \; -->}\times <!--\; \times \; -->u))=0$

If I have functions $f(x)=\sin <!--\; \; -->(x),q(x)=x^2+1\text{ and }s(x)=x^2+\sin <!--\; \; -->(x)$ which are linear combinations, how do I represent these functions as vectors? The sin(x) part is throwing me off when constructing the augmented matrix.
These functions are linear combinations of a vector space V of continuous functions. I want to find a basis of V. I don't understand how to construct the matrix in the appropriate way to find the basis vector.

Given d>1, $\mathbf{x}\in <!--\; \in \; -->{\mathrm{I}\mathrm{R}}^d$, and two multivariate functions $\mathbf{u},\mathbf{v}:{\mathrm{I}\mathrm{R}}^d\to <!--\; \to \; -->{\mathrm{I}\mathrm{R}}^d$, what is the second derivative of $f:{\mathrm{I}\mathrm{R}}^d\to <!--\; \to \; -->\mathrm{I}\mathrm{R}$
$f(\mathbf{x})=\mathbf{u}(\mathbf{x})\cdot <!--\; \cdot \; -->\mathbf{v}(\mathbf{x})=\mathbf{u}(\mathbf{x}{)}^T\mathbf{v}(\mathbf{x})$
with respect to x?
The first derivative is
$\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathbf{x}}(\mathbf{x})=\mathbf{u}(\mathbf{x}{)}^T\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{v}}{\mathrm{\partial <!--\; \partial \; -->}\mathbf{x}}(\mathbf{x})+\mathbf{v}(\mathbf{x}{)}^T\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{u}}{\mathrm{\partial <!--\; \partial \; -->}\mathbf{x}}(\mathbf{x})$
which is a row vector of size d when using the numerator layout.
But I do not know how to differentiate again since $\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{u}}{\mathrm{\partial <!--\; \partial \; -->}\mathbf{x}}$ and $\frac{\mathrm{\partial <!--\; \partial \; -->}\mathbf{v}}{\mathrm{\partial <!--\; \partial \; -->}\mathbf{x}}$ are 3-by-3 matrices...
I tried to develop these equations in a simple situation where d=2, but I obtain a quite complicated matrix and I am looking for a clear formula, in any dimension d>1, where only u,v and their derivatives appear.

$u=(1,1,0,2),v=(-<!--\; -\; -->1,0,3,4)$. Determine two perpendicular vectors a and b such that a is parallel to v and u=a+b
How can I solve this kind of problem? I tried testing with its cross product and trying to get $(u\times <!--\; \times \; -->v)\cdot <!--\; \cdot \; -->a=0$. But I think that's not correct.

Why is the vector v=i+j+4k perpendicular to the plane given by x+y+4z=0 ?
I understand that given an equation defining a plane, you can just read off the coefficients for each variable to get a normal vector.

In my notes, it says to consider an arc length parameterization r(s). Then we can show that B' points in the direction of N such that we can write $B^{\prime }=-<!--\; -\; -->𝜏N$. I understand why $||B^{\prime }||=𝜏$ but am unsure how to show why B' is parallel to N? Also, is the minus sign just for convention or is there a reason it is there?

Are two 2D vectors linearly dependent when $x_1{y}_2=x_2{y}_1$?
I was working on a proof for why a set of three vectors $\in <!--\; \in \; -->{\mathbb{R}}^2$ are always linearly dependant when I came along this.
It appears that given two vectors $[x_1,x_2{]}^T$ and $[y_1,y_2{]}^T$, if $x_1{y}_2=x_2{y}_1$ then one vector is a scalar multiple of the other. This is the case in my proof where you have to divide by zero (more accurately, divide by $x_1{y}_2-<!--\; -\; -->x_2{y}_1$) so I feel like this can't be a coincidence.
An example is when $x_1{y}_2=x_2{y}_1=12$, we could make pairs of vectors like $v_1=[12,4{]}^T,v_2=[3,1{]}^T$ and then $v_1=4v_2$
Does this hold for all vector pairs that satisfy the restriction and if so, why?

How can I show that
${\int <!--\; \int \; -->}_{\mathrm{\Sigma <!--\; \Sigma \; -->}}(\phi {\psi }_{,i}-<!--\; -\; -->\psi {\phi }_{,i})d^{2}x={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\phi {\psi }_{,ii}-<!--\; -\; -->\psi {\phi }_{,ii})d^{3}x$
, where $\psi ,\phi$ are scalar functions of the coordinates. Then i,ii designate differentiation of first and second order, and $\Sigma ,\Omega$ are the surface and volyme respectively?

Let x be a block vector with two vector elements, $x=(a,b)$ where a and b are vectors of size n and m, respectively.
Express $avg(x)$ in terms of $avg(a)$, $avg(b)$, m, and n.

I have a plane V with cartesian equation 2x−y+3z=10. I also have a line that is perpendicular to the plane V with parametric equation (x,y,z)=(1,0,−2)+(2,−1,3)t.
How do I find the point of intersection between the line and plane?

I am dealing with sets of vectors $\{x_1,x_2,x_3,\dots <!--\; \dots \; -->,x_m\}$ from some abstract vector space $\mathcal{V}$. Occasionally, $\mathcal{V}={\mathbb{R}}^n$ an I need to address the elements these vectors e.g. sum over all elements of the vector $x_j$. However, the subindex is already used to denote a specific vector.
Is there a common notation address the elements of a vector? E.g. $x_j[i]$ or $x_j^i$ or $x_j^{(i)}$?

Writing a vector a a sum of two vectors
Write $\mathbf{u}=\left[\begin{array}{c}9\\ -<!--\; -\; -->1\\ 2\end{array}\right]$ as a sum of two vectors where one is orthogonal to the plane
x+2y−2z=0 and the other is parallel to the same plane.
I've attempted to use the fact that the normal vector to the plane, $\left[\begin{array}{c}1\\ 2\\ -<!--\; -\; -->2\end{array}\right]$, can be seen from the given plane equation above. We should be able to write u as
$\mathbf{u}=\left[\begin{array}{c}9\\ -<!--\; -\; -->1\\ 2\end{array}\right]=s\left[\begin{array}{c}1\\ 2\\ -<!--\; -\; -->2\end{array}\right]-<!--\; -\; -->t\left[\begin{array}{c}a\\ b\\ c\end{array}\right].$
How do I find a,b,c in the vector that is parallel to the plane?

Given two lines:
$r_1(t)=(a_1,b_1,c_1)+t(p_1,q_1,r_1)$
$r_2(s)=(a_2,b_2,c_2)+s(p_2,q_2,r_2)$
I have calculated the cross product of the direction vectors to get the vector perpendicular to both lines:
$\stackrel{\to <!--\; \to \; -->}{n}=(p_1,q_1,r_1)\times <!--\; \times \; -->(p_2,q_2,r_2)$
Which gives the vector $\stackrel{\to <!--\; \to \; -->}{n}$ .
I then got a vector $\stackrel{\to <!--\; \to \; -->}{V}$ between a point on each line, and calculated the minimum distance d between both lines by finding the scalar projection of $\stackrel{\to <!--\; \to \; -->}{V}$ onto $\stackrel{\to <!--\; \to \; -->}{n}$
$d=\frac{|\stackrel{\to <!--\; \to \; -->}{V}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{n}|}{|\stackrel{\to <!--\; \to \; -->}{n}|}$
I now need the points on both lines where this minimum distance is achieved, how would I go about doing that?

Let A,B,C be the points of triangle and P inside of triangle. How can I prove using vectors that:
$(\stackrel{\to <!--\; \to \; -->}{AP}+\stackrel{\to <!--\; \to \; -->}{CP}+\stackrel{\to <!--\; \to \; -->}{BP})=\stackrel{\to <!--\; \to \; -->}{0}$
only if P is centroid

A line with equation $r=a+\mathrm{\lambda <!--\; \lambda \; -->}\stackrel{\to <!--\; \to \; -->}{d}$ meets plane π with equation $r.\stackrel{^<!--\; ^\; -->}{n}=k$ at point P. Point Q lies in π and is the foot of the perpendicular from A to π. Find the direction vector of line PQ.
By solving $(a+\mathrm{\lambda <!--\; \lambda \; -->}\stackrel{\to <!--\; \to \; -->}{d}).\stackrel{^<!--\; ^\; -->}{n}=k$, I was able to find the position vector of P. Then by finding the intersection of line AQ and plane I was able to find the position vector of Q and hence the direction vector PQ.
However, the answer can be found simply by finding $(\stackrel{^<!--\; ^\; -->}{n}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{d})\times <!--\; \times \; -->\stackrel{^<!--\; ^\; -->}{n}$ where × is cross-product. I don't understand why.
Here's what I know : The cross-product of 2 vectors gives a 3rd vector perpendicular to the 2 vectors. Line PQ lies on plane so direction vector PQ $\perp <!--\; \perp \; -->\stackrel{^<!--\; ^\; -->}{n}$. Also, AQ is parallel to $\stackrel{^<!--\; ^\; -->}{n}$.
The first part $w=(\stackrel{^<!--\; ^\; -->}{n}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{d})$ gives a vector perpendicular to line and parallel to plane. Won't $w\times <!--\; \times \; -->n$ give a vector perpendicular to the plane again? I can't understand the geometric interpretation of $(\stackrel{^<!--\; ^\; -->}{n}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{d})\times <!--\; \times \; -->\stackrel{^<!--\; ^\; -->}{n}$.

Prove that a parametric equation's range is subset of a cartesian equation
$r(t)=t(t-<!--\; -\; -->2{)}^{3}i+t(t-<!--\; -\; -->2{)}^{2}j$ where $r:\mathbb{R}\to <!--\; \to \; -->{\mathbb{R}}^2$
$C=\{(x,y)\in {\mathbb{R}}^2\mid <!--\; \mid \; -->y^4=x^3+2x^{2}y\}$
I have difficulty with this question. This is where I am up to.
Find $dy/dx$ of the cartesian curve $dy/dx=(3x^2+4xy)/(4y^3-<!--\; -\; -->2x^2)$
$(4y^3-<!--\; -\; -->2x^2)=0,y=(2x^{2/3})/2$
Substitute y into cartesian equation and find $x=-<!--\; -\; -->27/16,0$
I'm not sure where to go from here, can anyone clarify if I am on the right track, thanks!