Vector Examples, Equations, and Practice Problems

Suppose a cable car moves with direction vector $\stackrel{\to <!--\; \to \; -->}{d}$
$\left(\begin{array}{c}2\\ -<!--\; -\; -->2\\ 1\end{array}\right)$
The question asks, "at what angle to the horizontal does the cable car travel?". By horizontal, if you assume the x-axis, the angle can be found using
$\frac{|\stackrel{\to <!--\; \to \; -->}{a}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{b}|}{|\stackrel{\to <!--\; \to \; -->}{a}\stackrel{\to <!--\; \to \; -->}{b}|}$
) Using this, the angle I find between $\stackrel{\to <!--\; \to \; -->}{d}$and
$\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right)$
The angle I find is ${48.2}^{\circ <!--\; \circ \; -->}$ . If I use the z-vector instead:
$\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$
The angle I find is ${70.5}^{\circ <!--\; \circ \; -->}$
The answer is $\approx <!--\; \approx \; -->{19.5}^{\circ <!--\; \circ \; -->}$. How would I get this value?

Consider the vectors $u=5i+4j+3k$ , $v=-4i-j-k$ and $w=4i+4j-5k$ . Compute $|5u-<!--\; -\; -->4v+w|$

If $x_n,x,y$ are vectors in $R^m$ and $x_n\to <!--\; \to \; -->x$, prove that $⟨<!--\; ⟨\; -->x_n,y⟩<!--\; ⟩\; -->\to <!--\; \to \; -->⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->$
I am assuming one can simply say that $x_n$ $\to <!--\; \to \; -->$ x so we get the result but I don’t understand how vectors work and an explanation of how this proof works would go a long way.

If there are two vectors given
$\stackrel{\to <!--\; \to \; -->}{OA}=\stackrel{^<!--\; ^\; -->}{i}+2\stackrel{^<!--\; ^\; -->}{j}\text{and}\stackrel{\to <!--\; \to \; -->}{OB}=4\stackrel{^<!--\; ^\; -->}{i}+p\stackrel{^<!--\; ^\; -->}{k}$
Then find the values of p for which $\mathrm{\angle <!--\; \angle \; -->}AOB={\cos }^{-<!--\; -\; -->1}<!--\; \; -->(\frac{1}{5})$
Would someone please give me a hand on solving this?

How to solve the vector equation
Given the following vector equation in three dimensions
$\mathbf{r}+(\mathbf{r}\times <!--\; \times \; -->\mathbf{d})=\mathbf{c}$
where c and d are fixed given vectors, how can you find all solutions for r?
So far I have tried the following steps, to show that r must lie in a given plane.
By taking the dot product with d on both sides, we obtain
$\mathbf{r}+(\mathbf{r}\times <!--\; \times \; -->\mathbf{d})=\mathbf{c}⟹<!--\; ⟹\; -->(\mathbf{r}+(\mathbf{r}\times <!--\; \times \; -->\mathbf{d}))\cdot <!--\; \cdot \; -->\mathbf{d}=\mathbf{c}\cdot <!--\; \cdot \; -->\mathbf{d}.$
Since $\mathbf{r}\times <!--\; \times \; -->\mathbf{d}$ is perpendicular to d, their dot product is zero, so we get the following (Equation ∗)
$\mathbf{r}\cdot <!--\; \cdot \; -->\mathbf{d}=\mathbf{c}\cdot <!--\; \cdot \; -->\mathbf{d}$
from which we can deduce that r (as a position vector) lies in the plane that contains c and is normal to the vector d.
However this doesn't necessarily imply that all points in this plane are valid solutions for r. I can't see how Equation (∗) can be substituted back into the original equation to somehow eliminate a term in r or simplify it. How do you solve this equation, making sure that you find all solutions for r?

I am having trouble finding a vector that is perpendicular to both $\stackrel{\to <!--\; \to \; -->}{a}=-<!--\; -\; -->9\stackrel{^<!--\; ^\; -->}{i}+2\stackrel{^<!--\; ^\; -->}{j}+6\stackrel{^<!--\; ^\; -->}{k}$ and $\stackrel{\to <!--\; \to \; -->}{b}=-<!--\; -\; -->2\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}+\stackrel{^<!--\; ^\; -->}{k}$
Would it just be $18i+2j+6k?$

I have been trying to solve this exercice and I would like to know if my solution is correct or not ?
Point is $P=(1,7,4)$
Plane has an equation : $5x+3y+z=8$
I have taken a random point on the plane Q=(1,1,0)
$\stackrel{\to <!--\; \to \; -->}{PQ}=<0,-<!--\; -\; -->6,-<!--\; -\; -->4>$
Normal to the plane is :
$\stackrel{\to <!--\; \to \; -->}{N}=<5,3,1>$
(by reading the coefficients of the plane's equation)
$Distance=|\stackrel{\to <!--\; \to \; -->}{PQ}\ast <!--\; \ast \; -->\frac{\stackrel{\to <!--\; \to \; -->}{N}}{|<!--\; |\; -->N|<!--\; |\; -->}|=\frac{0-<!--\; -\; -->18-<!--\; -\; -->4}{\sqrt{5²+3²+1²}}=\frac{22}{\sqrt{35}}$

Let u,v be vectors such that $||u||=2,||v||=\sqrt{3}$ , and $u\cdot <!--\; \cdot \; -->v=1$. Find $||u+v||$
So far I calculated ${\displaystyle \frac{u\cdot <!--\; \cdot \; -->v}{||u||\ast <!--\; \ast \; -->||v||}}=\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ and then with one angle and two sides in the triangle known, I can calculate the length of (u−v) as the third side of the triangle.

S is a vector subspace of vector space V? (matrix in complex)
$V=M_2(\mathbb{C})$ , K=C and
$S=\{\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\in <!--\; \in \; -->V:a_{ij}=\stackrel{¯<!--\; ¯\; -->}{a_{ji}}\text{ for }i,j=1,2\}.$
I'd like to understand better how to work with complex and matrix to check this question. I know there are 3 rules, but id like how to apply in cases like that. Thanks in advance, ill be glad for explanations.

Proving that $\frac{\text{d}\stackrel{\to <!--\; \to \; -->}{v}}{\text{dt}}=\frac{\text{d}|\stackrel{\to <!--\; \to \; -->}{v}|}{\text{dt}}\stackrel{^<!--\; ^\; -->}{v}+\frac{\text{d}\stackrel{^<!--\; ^\; -->}{v}}{\text{dt}}|\stackrel{\to <!--\; \to \; -->}{v}|$
where $\stackrel{\to <!--\; \to \; -->}{v}$ is a velocity vector dependent on time t. I was wondering how the equation could be proven.

Let $\Vert <!--\; \Vert \; -->a-<!--\; -\; -->b\Vert <!--\; \Vert \; --><\Vert <!--\; \Vert \; -->b\Vert <!--\; \Vert \; -->$ where $a,b\in <!--\; \in \; -->C^n$. Аre they following any relations for $\Vert <!--\; \Vert \; -->a\Vert <!--\; \Vert \; -->$ or $\Vert <!--\; \Vert \; -->b\Vert <!--\; \Vert \; -->$?

I will be so grateful if someone can explain me in a simple language what is difference between vector magnitude namely
$r=|{\stackrel{\to <!--\; \to \; -->}{r}}_1-<!--\; -\; -->{\stackrel{\to <!--\; \to \; -->}{r}}_2|$
and distance vector namely
${\stackrel{\to <!--\; \to \; -->}{r}}_{12}={\stackrel{\to <!--\; \to \; -->}{r}}_1-<!--\; -\; -->{\stackrel{\to <!--\; \to \; -->}{r}}_2$

Suppose ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$ are real vectors of length $N>3$
If we use Gram-Schmidt process, we can find two orthogonal vectors ${\mathbf{u}}_1,{\mathbf{u}}_2$ such that
${\mathbf{u}}_1={\mathbf{v}}_1,$
${\mathbf{u}}_2={\mathbf{v}}_2-<!--\; -\; -->\frac{<{\mathbf{v}}_2,{\mathbf{u}}_1>}{<{\mathbf{u}}_1,{\mathbf{u}}_1>}{\mathbf{u}}_1,$
where $<\mathbf{x},\mathbf{y}>$ denotes the inner product between the two vectors $\mathbf{x},\mathbf{y}$
But I am wondering whether it is possible to find a vector z that is mutually orthogonal to ${\mathbf{v}}_1,{\mathbf{v}}_2$

For what t will the following vector be an orthogonal basis?
$\begin{array}{rl}{u}_1& =(1,t,t)\\ u_2& =(2t,t+1,2t-<!--\; -\; -->1)\\ u_3& =(2-<!--\; -\; -->2t,t-<!--\; -\; -->1,1)\end{array}$
Till now I have tried using the Gram-Schmidt process but did not really reach anywhere. Can you please provide a hint or some theory that may help me get the solution for this question?

I would like to ask how to prove these two formulas.
for T is a scalar and r is radial component of a vector in spherical coordinate.
(1) $\mathbf{r}\cdot <!--\; \cdot \; -->\mathbf{[}\mathbf{(}\mathbf{r}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{)}\mathrm{\nabla <!--\; \nabla \; -->}T+2\mathrm{\nabla <!--\; \nabla \; -->}T]=\mathbf{r}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}T+\mathbf{r}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{(}\mathbf{r}\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}T)$
(2) ${\mathrm{\nabla <!--\; \nabla \; -->}}^2(T\mathbf{r}\mathbf{)}=\mathbf{r}{\mathrm{\nabla <!--\; \nabla \; -->}}^{2}T+T{\mathrm{\nabla <!--\; \nabla \; -->}}^2\mathbf{r}+2\mathrm{\nabla <!--\; \nabla \; -->}T\cdot <!--\; \cdot \; -->\mathrm{\nabla <!--\; \nabla \; -->}\mathbf{r}$

How come ${\mathrm{\nabla <!--\; \nabla \; -->}}^{2}f=\frac{1}{r^2}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}r}(r^2\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}r})$ yet $f=\int <!--\; \int \; -->\int <!--\; \int \; -->f^{″}drdr$?
In spherical coordinate, suppose f dependents only on r. Then ${\mathrm{\nabla <!--\; \nabla \; -->}}^{2}f=\frac{1}{r^2}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}r}(r^2\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}r})$. However, to integral f′′ back the f, $\int <!--\; \int \; -->\int <!--\; \int \; -->f^{″}drdr$
How could that be?

I have a 3D vector with length != 1 but all components lie within [-1, 1].
now I want to create a unit vector such that this unit vector has the same x coordinate as the original vector and the cross product between the x-axis and the new vector shall point in the same direction as the cross product between the x-axis and the original vector.
How do I need to scale the y and z coordinates to achieve this?

Find the equation of the plane that contains the line $x=1+t,y=2-<!--\; -\; -->t,z=4-<!--\; -\; -->3t$ and is parallel to the plane $5x+2y+z=1$
So far I figured the direction vector of the line is $<1,-<!--\; -\; -->1,-<!--\; -\; -->3>$ and that the symmetric equation of the line is $x-<!--\; -\; -->1=2-<!--\; -\; -->y=\frac{z-<!--\; -\; -->4}{-<!--\; -\; -->3}$. But I am lost on how to go from here.

What is the distance between the line with Cartesian equation x−4=y−5=z and the line with Cartesian equation x−2=y−5=z−1?
I am not sure how to do it. I think the lines are parallel with the same direction but I'm not sure.

Let $A_1,A_2,...,A_r$ Some non-zero vectors in ${\mathbb{R}}^n$ such that $A_i\cdot <!--\; \cdot \; -->A_j=0$ for all $i\ne <!--\; \ne \; -->j$
Let $c,...,c_r$ some real numbers such that
$c_1{A}_1+c_2{A}_2+\cdots <!--\; \cdots \; -->+c_r{A}_r=0$
Prove that all $c_i=0$
First it is more simple to work with the special case r=2, then since $A_1\cdot <!--\; \cdot \; -->A_2=0$ Then there is no such $c_1,c_2\in <!--\; \in \; -->{\mathbb{R}}^2$ such that
$c_1{A}_1+c_2{A}_2=0$
except for the case where $c_1=c_2=0$
This last statement has to formalize in a more rigorous way, but i don’t how to do it i’ve just an intuition that this is the right case, and for higher dimensional vector this intuition would disappear.