Learn Vectors with Examples, Equations, and Practice Problems

A linear map $f : R^{3} \to R^{3}$ is defined as $\vec{u} \times (\vec{r} \times \vec{u})$ where $\vec{u}$ is a unit vector. What would its geometry look like?
I know that I can rewrite this map as $(\vec{u} \cdot \vec{u}) \vec{r} - (\vec{u} \cdot \vec{r}) \vec{u} = \vec{r} - (\vec{u} \cdot \vec{r}) \vec{u}$
However I am not sure what to do from here.

Javion Henry 2022-07-21

Given that $| \vec{n} | = 1, | \vec{m} | = \sqrt{3}$ And that the angle between the vectors $| \vec{m} |$ and $| \vec{n} |$ is $30^{\circ}$
We will define $\vec{a} = \vec{m} - \vec{n}, \vec{b} = \vec{m} + \vec{n}$ calculate the area of the triangle created by vectors $\vec{a}, \vec{b}$
I found out what $| \vec{a} |$ and $| \vec{b} |$ but not sure how to continue form here.
$\begin{aligned} {| \vec{a} |}^{2} & = (\vec{m} - \vec{n}) (\vec{m} - \vec{n}) \\ = {| \vec{m} |}^{2} - 2 \vec{m} \cdot \vec{n} + {| \vec{n} |}^{2} \\ = 3 - 2 \sqrt{3} \cdot 1 \cdot \frac{\sqrt{3}}{2} + 1 \\ = 1 ⟹ | \vec{a} | = 1 \end{aligned}$
$\begin{aligned} {| \vec{b} |}^{2} & = (\vec{m} + \vec{n}) \cdot (\vec{m} + \vec{n}) \\ = {| \vec{m} |}^{2} + 2 \vec{m} \cdot \vec{n} + {| \vec{n} |}^{2} \\ = 3 + 2 \sqrt{3} \cdot 1 \cdot \frac{\sqrt{3}}{2} + 1 \\ = 7 ⟹ | \vec{b} | = \sqrt{7} \end{aligned}$

Mauricio Mathis 2022-07-21

I'm having difficulty proving the formula:
$u \times ω = \nabla (\frac{u \cdot u}{2}) - u \cdot \nabla u$
I should be using tensor notation. Given is that:
$ω = \nabla \times u$
and
$\nabla \cdot u = 0$
I've done this so far:
$(u \times ω)_{i} = (u \times (\nabla \times u))_{i} = ϵ_{i j k} u_{j} (ϵ_{k l m} \frac{\partial}{\partial x_{l}} u_{m}) = ϵ_{i j k} ϵ_{k l m} u_{j} \frac{\partial}{\partial x_{l}} u_{m} = ϵ_{k i j} ϵ_{k l m} u_{j} \frac{\partial}{\partial x_{l}} u_{m} = (δ_{i l} δ_{j m} - δ_{i m} δ_{j l}) u_{j} \frac{\partial}{\partial x_{l}} u_{m} = u_{j} \frac{\partial}{\partial x_{i}} u_{j} - u_{j} \frac{\partial}{\partial x_{j}} u_{i}$
But that is as far as I come.

Ibrahim Rosales 2022-07-21

Two unit vectors x and y in $R^{n}$ satisfy $x \cdot y = \frac{\sqrt{2}}{2}$ in radians. How would I go about finding the angle between x and y?
As I don't know the x and y unit vectors, would the unit circle be useful here? For instance, using $\frac{\sqrt{2}}{2}$ and plugging those values into $\frac{x \cdot y}{‖ x ‖ ‖ y ‖}$ to find the angle?

Zoagliaj 2022-07-21

Vector parametrization of the line that passes through two vectors.
How would I find the vector parametrization $𝐫 (𝑡)$ of the line $L$ that passes through the points $(2, 1, 4)$ and $(5, 6, 7)$ ?
So I found a directional vector: $(3, 5, 3)$
What do I do next?

on2t1inf8b 2022-07-21

Given the equation:
$(2, - 14, 8) x + (- 3, 21, - 12) y = \vec{0}$
Find the scalars x and y, not both zero, such that the above equation holds.
My attempt:
I tried solving the equation system 2x−3y=0,−14x+21y=0,8x−12y=0 but I only obtained trivial solutions, i.e. x=0,y=0

Patricia Bean 2022-07-20

Show that vectors are in an affine plane if and only if det=0
Let $K$ be a field and, $a = (\begin{array}{l} a_{1} \\ a_{2} \\ a_{3} \end{array}), b = (\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}), c = (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix}), d = (\begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}) \in K^{3}$
Show that a,b,c,d are in an affine plane if and only if
$d e t (\begin{array}{llll} a_{1} & b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} & d_{3} \\ 1 & 1 & 1 & 1 \end{array}) = 0$
How can I show this?

on2t1inf8b 2022-07-20

The original problem is:
If a, b, c, d are the position vectors of points A, B, C, D respectively such that
$(\vec{a} - \vec{d}) . (\vec{b} - \vec{c}) = (\vec{b} - \vec{d}) . (\vec{c} - \vec{a}) = 0$
then prove that D is the orthocentre of $Δ$ ABC.
How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.

phepafalowl 2022-07-20

$[\begin{matrix} - 2 & 11 \\ 4 & 2 \end{matrix}]$ represents a linear transformation T: $R^{2}$ to $R^{2}$ with respect to the basis, [(3,1),(0,2)]. Find the matrix of T with respect to basis [(1,1),(−1,1)]

anudoneddbv 2022-07-20

How to find a vector given the scalar triple product
Scalar triple product formula: $a (b \times c)$
If I know the value of the product and I know what vectors b and c are, how do I find what vector a is equal to?

stratsticks57jl 2022-07-20

I want to prove
$| \vec{r} - \vec{s} |^{3} = (r^{2} + s^{2} - 2 \cdot r \cdot s \cdot \cos (θ))^{3 / 2}$
where $r = | \vec{r} |$ . but I have problems because
$(r^{2} + s^{2} - 2 \cdot r \cdot s \cdot \cos (θ))^{3 / 2} = (r^{2} + s^{2} - \vec{r} \cdot \vec{s})^{3 / 2} = ((\vec{r} - \vec{s})^{2})^{3 / 2} = (\vec{r} - \vec{s})^{3}$
I hope someone can help me.

Ashlyn Krause 2022-07-20

I have a vector field $\vec{u} = (x, z^{2}, y)$ over a tetrahedron with vertices at $(0, 0, 0), (1, 0, 0), (0, 1, 0)$ and $(0, 0, 2)$ and I need to compute $\int_{S} \vec{u} \cdot d \vec{S}$ being the surface of the tetrahedron.
I have already calculated $\int_{V} (\nabla \cdot \vec{u}) d V$ , which I have found to be equal to $\frac{1}{3}$ , now I need to calculate the previous integral and assumedly show that the answers agree (Gauss' Theorem).
So far I have parameterised the surface to say ${\vec{x}}_{s} = (s, t, 2 - 2 s - 2 t)$
Next I find $d \vec{S}$ as such:
$d \vec{S} = (\frac{\partial \vec{x_{S}}}{\partial s} \times \frac{\partial \vec{x_{S}}}{\partial t}) d s d t$
Therefore: $\vec{u} \cdot d \vec{S} = 10 s^{2} + 16 s t - 16 s + 8 t^{2} - 15 t + 8$
But, upon calculating the integral $\int_{0}^{1} \int_{0}^{1} \vec{u} \cdot d \vec{S} d s d t$ I get $\frac{5}{2}$ . I'm confident in my first integral, but I don't feel as confident in this one as I feel there may be some theoretical misunderstanding with my parameterisation. Any help is appreciated.

Brenton Dixon 2022-07-20

Show that the planes $a x + a z = c$ and $b x - b y = d$ , where $a, b, c, d \in R, a, b \neq 0$ , always form an angle of $π / 3 = 60^{\circ}$

Lisa Hardin 2022-07-20

I want to find the vertical or you could say perpendicular component of $\vec{a}$ on $\vec{b}$
Now I know that it can be found out using $\vec{a} - (\frac{a \cdot b}{| b |}) \vec{b}$
However I wanted to know why it cannot be found out using this method I tried. What is the flaw in it?
$\vec{a} \times \vec{b} = | a | | b | \sin θ \hat{n}$ Now $\hat{n}$ should be equal to $\frac{(\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |}$
Using this I can write the expression as $\vec{a} \times \vec{b} = | a | | b | \sin θ \times$ $\frac{(\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |}$ which on simplifying gives me
$\frac{| \vec{a} \times \vec{b} |}{| b |} = | a | \sin θ$ which I believe should be the perpendicular component
Now I'm probably doing something really stupid but I can't really understand where am I going wrong ?

comAttitRize8 2022-07-20

How to determine the value of x such that the parallelogram has a given area?
Determine the value(s) of x such that the area of the parallelogram formed by the vectors $a = (x + 1, 1, - 2)$ and $b = (x, 3, 0)$ is $\sqrt{41}$
My work (using cross product)
$\begin{aligned} | a \times b | & = ((1) (0) - (- 2) (3), (- 2) (x) - (x + 1) (0), (x + 1) (3) - (1) (x)) \\ | a \times b | & = (6, - 2 x, 3 x + 3 - x) \\ | a \times b | & = (6, - 2 x, 2 x + 3) \\ \sqrt{41} & = (6, - 2 x, 2 x + 3) \end{aligned}$
I don't know what to do next

Kade Reese 2022-07-20

Can I simplify the following vector?
Let a and b be $n \times 1$ vectors and $M_{1}, M_{2}, . . ., M_{n}$ be $n \times n$ matrices.
Is there a way to write the following vector:
$v = [\begin{matrix} a^{T} M_{1} b \\ a^{T} M_{2} b \\ ⋮ \\ a^{T} M_{n} b \end{matrix}]$
in such a way that reads $v = a^{T} M_{n e w} b$ , where $M_{n e w}$ is just a manipulation of the $M_{i}$ matrices?
I know I can do:
$v = [\begin{matrix} a^{T} M_{1} \\ a^{T} M_{2} \\ ⋮ \\ a^{T} M_{n} \end{matrix}] b$
But then I cannot get the vector a "out of the brackets".

Makena Preston 2022-07-20

This is the question. The vectors u and v are such that $‖ u ‖ = 4$ , $‖ v ‖ = 5$ and $(u | v) = - 12$ . Compute the length of the vector $(2 u + 3 v) \times (5 v - 4 u)$
I broke it down into $2 u \times (5 v - 4 u) + 3 v (5 v - 4 u)$ and then further broke it down into $(2 u \times 5 v) - (2 u \times 4 u) + (3 v \times 5 v) - (3 v \times 4 u)$
But what I don't get is, how do I use $(u | v) = - 12$ here in order to compute the answer? Is it connected to the formula $u \times v = ‖ v ‖ T (u^{'})$ ?

Braylon Lester 2022-07-20

Determine vector c, which is collinear vector of vector a+b, if ab=5, cb=18 and |b|=2.
I tried with $c = n (a + b)$
$9 = | c | * c o s (α)$ ... $| c | = \sqrt{(n^{2} (a + b) (a + b))} = n \sqrt{a a + 14}$
Then $9 = n \sqrt{a a + 14} * c o s (α)$
Second equation is: $\frac{5}{2} \sqrt{a a} * c o s (α)$
From second equation we get: $c o s (α) = \frac{5}{2 \sqrt{a a}}$ . I put this in first equation and I get that $n = \frac{9 \sqrt{14}}{35}$
My solution is: $c = \frac{9 \sqrt{14}}{35} (a + b)$
Is this correct?

Alonzo Odom 2022-07-20

Let k be an infinite field.
Let V be a vector space over k and $W_{1}, . . ., W_{r}$ proper subspaces of V.
Show that $⋃_{i = 1}^{r} W_{i} \neq V .$
I tried the following:
for all $j \in {1, . . ., r}$ , I take $w_{j} \in W_{j}$ such that $w_{j} \notin W_{i}$ whenever $j \neq i$ , so I know that $w_{1} + \dots + w_{r} \in V$ . If $w_{1} + \dots + w_{r} \in ⋃_{i = 1}^{r} W_{i}$ , then there is $l \in {1, . . ., r}$ such that $w_{1} + \dots + w_{r} \in W_{l}$ . I don't find because $w_{1} + \dots + w_{r} \in W_{l}$ is absurd.
Is this correct reasoning, or is there other way for me to prove this?

Luciano Webster 2022-07-19

$a (\begin{matrix} 1 \\ 4 \end{matrix}) + b (\begin{matrix} 3 \\ - 1 \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix})$ Find a and b
I've found the answers to this to be $\frac{1}{13}$ and $\frac{4}{13}$ through google but just wondering, how would I go about working this through and finding the answer myself?

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