Learn Vectors with Examples, Equations, and Practice Problems

Show that $\nabla (v \cdot A) - (v \cdot \nabla) A = v \times (\nabla \times A)$
All of the variables are vectors. It is based on the lagrangian of a electromagnetic field, where v is the velocity and A the cross potential.
I've tried to prove it by using their modulus, sines and cosines but I don't think that's the way to prove it.
Could you give me any hint?

Juan Lowe 2022-11-14

Prove the identity between vectors A, B, C
$2 (A . B) (A . C) = (A . A) (B . C) + A^{2} . B C$
$A^{2}$ is neither a dot or cross product

Bayobusalue 2022-11-13

$U_{x}$ will be equal to $u \cos θ$ and $u = \sin θ$ . But in case of final velocity, it is $v \cos θ$ where $- g t = 0$ and $v \sin θ - g t$
Why is that ? I am not getting how to solve this.

figoveck38 2022-11-13

Let $θ_{0}$ be a real number. Define $R : R^{2} \to R^{2}$ by $R (x, y) = ((c o s θ_{0}) x - (s i n θ_{0}) y, (s i n θ_{0}) x + (c o s θ_{0}) y)$
How do I show that R rotates both i and j by the angle $θ_{0}$ counterclockwise? I know that R should be a linear transformation

Emmanuel Giles 2022-11-11

Given $A = {v | \forall i \in N N : v_{i} \in {0, 1} and {v_{i} = 1}} is finite group}$ , namely, That is, all infinite binary vectors so that they have a finite number of "1".Someone have ideas how can I define injective function $f : A \to N$ ?

Paula Cameron 2022-11-11

Let $\vec{a}, \vec{b}, \vec{c}$ be non-complanar vectors so that $\vec{a} \times \vec{b}, \vec{b}, \vec{c} \times \vec{a}$ on the same plane. How can I show that $\vec{a}$ is perpendicular to $\vec{b}$ ?

vidamuhae 2022-11-11

Can someone clarify this for me:
Suppose we have two column vectors: A = $[\begin{matrix} a \\ b \\ c \end{matrix}]$ and, B = $[\begin{matrix} x \\ y \\ z \end{matrix}]$
If M consists of A & B as rows, i.e, M = $[\begin{matrix} a & b & c \\ x & y & z \end{matrix}]$
Then, is it correct to write M = $[\begin{matrix} A^{⊤} \\ B^{⊤} \end{matrix}]$ ?
I have seen people usually write it directly as M = $[\begin{matrix} A \\ B \end{matrix}]$ . Here the transpose symbol is omitted. Isn't this incorrect?
We are basically transposing the original vectors when writing them in the matrix, therefore the first representation should be correct.
Any help is appreciated.

Ty Gaines 2022-11-11

Find the Cartesian equation corresponding to the following vector equation:
$r (t) = \frac{(e^{t} + e^{- t})}{2} \hat{i} + + \frac{(e^{t} - e^{- t})}{2} \hat{j}$

atgnybo4fq 2022-11-11

Suppose A and B are two non-parallel vectors. Where A and B are given as $A = (a_{1}, a_{2}, a_{3}, . . . . . a_{n})$ and $B = (b_{1}, b_{2}, b_{3}, . . . . . b_{n})$ How can I geometrically determine the angle $θ$ between the two vectors A and B? I know that $A . B = ‖ A ‖ ‖ B ‖ \cos (θ)$

Alvin Parks 2022-11-11

Determining divergence and gradient using chain rule
Suppose $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ and $r = | r |$ . Also, let $f$ be a scalar function of r and let $\vec{A}$ be a vector function of r. We wish to determine $\nabla f$ and $\nabla . \vec{A}$ .
It seems to be obvious that the chain rule is the way to go here:
$\nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}$
$= \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} \hat{i} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} \hat{j} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial z} \hat{k}$
$= \frac{\partial f}{\partial r} (\frac{\partial r}{\partial x} \hat{i} + \frac{\partial r}{\partial y} \hat{j} + \frac{\partial r}{\partial z} \hat{k})$
$= \frac{\partial f}{\partial r} \hat{r}$
Which is correct. However,
$\nabla . \vec{A} = \frac{\partial \vec{A}}{\partial x} . \hat{i} + \frac{\partial \vec{A}}{\partial y} . \hat{j} + \frac{\partial \vec{A}}{\partial z} . \hat{k}$
$= \frac{\partial \vec{A}}{\partial r} \frac{\partial r}{\partial x} . \hat{i} + \frac{\partial \vec{A}}{\partial r} \frac{\partial r}{\partial y} . \hat{j} + \frac{\partial \vec{A}}{\partial r} \frac{\partial r}{\partial z} . \hat{k}$
$= \frac{\partial \vec{A}}{\partial r} . (\frac{\partial r}{\partial x} \hat{i} + \frac{\partial r}{\partial y} \hat{j} + \frac{\partial r}{\partial z} \hat{k})$
$= \frac{\partial \vec{A}}{\partial r} . \hat{r}$
This expressions turns out to be incorrect.
The only scope for error seems to be in the usage of the chain rule, however if it(my usage of the chain rule) worked for the gradient, why did it fail here ?

Hallie Stanton 2022-11-11

I want to prove that if matrices $E_{i, j}$ in $K^{m, n}$ are matrices with all zeros except for one 1 on the $i, j$ index, then $tr ((E_{i, j})^{T} E_{k, l})$ equals 1 if $E_{i, j} = E_{k, l}$ and 0 in other cases. I have an intuition that this is true but I don't know how to formally prove it.

Audrey Arnold 2022-11-11

Say we have an orthonormal basis in $R^{3}$ {A,B,C}. Then when taking the cross product of any 2 of these, we know it is equal to either the third basis element, or −1 times that element.
Say we are given $A \times B = C$ , and we want to know whether $A \times C = B$ or $A \times C = - B$
Via the right-hand rule, I can see that it should be −B, but how can I show this algebraically, using only what is given here?

Owen Mathis 2022-11-11

The minimum distance of origin from the locus of point satisfying the equation $\vec{r} \times \vec{a} = \vec{b}$ $\vec{b} \neq 0$ is
(A) $\frac{| \vec{a} . \vec{b} |}{a^{2}}$
(B) $| \frac{\vec{a} + \vec{a} \times \vec{b}}{a^{2}} |$
(C) $| \frac{\vec{a} - \vec{a} \times \vec{b}}{a^{2}} |$
(D) $\frac{| \vec{a} \times \vec{b} |}{a^{2}}$
My approach is as follow
$\vec{r} \times \vec{a} = \vec{b} \Rightarrow \vec{a} \times (\vec{r} \times \vec{a}) = \vec{a} \times \vec{b} \Rightarrow (\vec{a} . \vec{a}) \vec{r} - (\vec{a} . \vec{r}) \vec{a} = \vec{a} \times \vec{b}$
How will I proceed from here and what condition I need to use so that the locus of point is minimum.

unabuenanuevasld 2022-11-10

If $\hat{x}$ , $\hat{y}$ and $\hat{z}$ are three unit vectors in three-dimensional space, then the minimum value of $| \hat{x} + \hat{y} |^{2} + | \hat{y} + \hat{z} |^{2} + | \hat{z} + \hat{x} |^{2}$ is:
My Attempt : Let $θ_{1}$ be angle between $\hat{x}$ and $\hat{y}$ , $θ_{2}$ be angle between $\hat{y}$ and $\hat{z}$ and $θ_{3}$ be angle between $\hat{z}$ and $\hat{x}$
$∴ θ_{1}, θ_{2}$ and $θ_{3}$ are the angles between any 2 edges of a tetrahedron from a single vertex.
$∴ | \hat{x} + \hat{y} |^{2} + | \hat{y} + \hat{z} |^{2} + | \hat{z} + \hat{x} |^{2} = (2 + 2 \cos θ_{1}) + (2 + 2 \cos θ_{2}) + (2 + 2 \cos θ_{3})$
$= 6 + 2 (\cos θ_{1} + \cos θ_{2} + \cos θ_{3})$
Can anyone please tell me what to do next?

Aron Heath 2022-11-09

Prove this generalization of Cauchy-Schwarz: $| (a . b) c + (b . c) a - (c . a) b | \leq | a | | b | | c |$
If you take c unit perpendicular to a,b, Cauchy-Schwarz follows.
I can prove the inequality in a roundabout way. First normalize to unit vectors. Then show that equality holds for 2-dimensional vectors (a bit messy). Then split c=d+e with $d \in S p a n (a, b)$ , $e ⊥ d$ and apply
$| (a . b) d + (b . d) a - (a . d) b + (a . b) e |^{2} = | d |^{2} + (a . b)^{2} | e |^{2} \leq | d |^{2} + | e |^{2} = 1$
But it's begging for a straightforward proof.

MISA6zh 2022-11-09

Find angle between the vector $\vec{a} =< 1, 2, 2 >$ and $\vec{b} =< 3, 4, 0 >$ in nearest degree
What I try:
$\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |} = \frac{3 + 8}{3 \times 5} = \frac{11}{25}$
I did not understand how can I convert it into degrees.

Paula Cameron 2022-11-09

I have the following question. In physics forces are vectors. Now I may write a force as
$F = F e_{F}$
with F denoting the length and $e_{F}$ denoting the direction vector. But some forces are the result of a cross product (pseudo vectors). The length is then
$| a | | b | \sin (θ)$
with a,b some vectors (maybe position and velocity) and $θ$ the angle between them. However these vectors are also sometimes written in the first form. How can I check whether F in the first form is the length of a cross product, or not?

Uriah Molina 2022-11-08

Find the equation of a plane that goes through the intersection of the planes $x_{1} + 5 x_{2} + x_{3} - 5 = 0$ and $x_{1} - 3 x_{3} - 2 x_{3} = 0$ and contains the point (1,1,0)
I calculated the cross product of the given plane to obtain the intersection and it gave me $- 7 x_{1} + 3 x_{2} - 8 x_{3}$ , but I'm stuck there.

evitagimm9h 2022-11-08

I have two vectors $v, w \in R^{n}$ . Is the statement $v^{T} w = w^{T} v$ true?
In my opinion yes, because both are a single number, and this should be the same.
Let $v := (\begin{matrix} x \\ y \\ z \end{matrix})$ and $w := (\begin{matrix} a \\ b \\ c \end{matrix})$
Then we get
$v^{T} w = (\begin{matrix} x & y & z \end{matrix}) (\begin{matrix} a \\ b \\ c \end{matrix}) = x a + y b + z c = a x + b y + c z = (\begin{matrix} a & b & c \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = w^{T} v$
because the multiplication is commutative.
Is this ok?

Davirnoilc 2022-11-08

Find the internal angle of the triangle between $\vec{a} & \vec{b}$
If $\vec{a} = \hat{j} + \sqrt{3} \hat{k}; \vec{b} = - \hat{j} + \sqrt{3} \hat{k}; \vec{c} = 2 \sqrt{3} \hat{k}$ form a triangle then find the internal angle of the triangle between $\vec{a} & \vec{b}$
My approach is as follow
Let $\vec{a} + \vec{b} = \vec{c}$
$\vec{a} = \hat{j} + \sqrt{3} \hat{k}; \vec{b} = - \hat{j} + \sqrt{3} \hat{k}; \vec{c} = 2 \sqrt{3} \hat{k}$
$\Rightarrow {(\vec{a} + \vec{b})}^{2} = {(\vec{c})}^{2} \Rightarrow {\vec{a}}^{2} + {\vec{b}}^{2} + 2 \vec{a} . \vec{b} = {\vec{c}}^{2}$
$\Rightarrow \vec{a} . \vec{b} = 2 \Rightarrow$
$\cos θ = \frac{\vec{a} . \vec{b}}{| \vec{a} | . | \vec{b} |} = \frac{1}{2} = \frac{π}{3}$
But official answer is $\frac{2 π}{3}$ where I am making mistake

Vectors in the Precalculus course are usually more challenging since there are different vectors examples that are always mentioned. For example, if you are majoring in Engineering disciplines, you will have to use more than one approach to explain the most efficient ways. Take a look at vectors practice problems that have been presented below. It will help you learn and find the answers that will let you see the best equation and graphs. At the same time, when you are dealing with vectors equations, do not forget about unknown coefficients as you are looking for combinations and various solutions.

Precalculus

Vector Examples, Equations, and Practice Problems