Learn Vectors with Examples, Equations, and Practice Problems

How can I prove that:
$\begin{matrix} (1) & 0 < r^{2} < (r^{2} + (w l)^{2}) ((1 - w^{2} l c)^{2} + (w r c)^{2}) \end{matrix}$
$\forall c > 0$ and all the other variables are bigger than zero using the scalar product of the vectors $A = (r, w l)$ and $B = (1 - w^{2} l c, w r c)$ ?
I do not know how to get started on this problem and I do not see how I can use the scalar product of vectors to prove this inequality.

Heergerneuu 2022-09-16

How to determine the span of two vectors in $R^{2}$
(4,2) and (1,3)
Do I subtract them? I don't how I'd solve this. Thanks in advance. In my question the vectors are like this:
$[\begin{matrix} 4 \\ 2 \end{matrix}]$
But that doesn't matter, right?
Would the vector equation $x_{1} v_{1} + x_{2} v_{2}$ = b be consistent for any b in $R^{2}$ ?

Zack Chase 2022-09-16

In my syllabus we have the alternative definition of the condition of a matrix:
$κ (A) = \frac{{max}_{‖ \vec{y} ‖ = 1} ‖ A \vec{y} ‖}{{min}_{‖ \vec{y} ‖ = 1} ‖ A \vec{y} ‖}$
In it, it also says that by definition of the condition of a matrix it follows that $κ (A^{- 1}) = κ (A)$ . So there is no explanation for this. Therefore, my question is: Why is $κ (A^{- 1}) = κ (A)$

trkalo84 2022-09-15

We have linearly independent vectors $e_{1}, e_{2}, . . ., e_{m + 1} \in R^{n}$ . I would like to prove that among their linear combinations, there is a nonzero vector whose first m coordinates are zero.
From independence we can get that $m + 1 \leq n$ . Maybe we can build some system, but what next?

malaana5k 2022-09-14

Given a cross product:
$\vec{u} \times \vec{v} = ⟨ - 1, 1, - 3 ⟩$
I'm trying to find:
$(\vec{u} - 3 \vec{v}) \times (\vec{u} + 2 \vec{v})$ as a vector.

Krish Crosby 2022-09-14

If p,q,r,x,y,z are non zero real number such that
$p x + q y + r z + \sqrt{(p^{2} + q^{2} + r^{2}) (x^{2} + y^{2} + z^{2})} = 0$
Then $\frac{p y}{q x} + \frac{q z}{r y} + \frac{r x}{p z}$ is
what try
$(p x + q y + r z)^{2} = (p^{2} + q^{2} + r^{2}) (x^{2} + y^{2} + z^{2})$
$p^{2} x^{2} + q^{2} y^{2} + r^{2} z^{2} + 2 p q x y + 2 q r y z + 2 p r x z = p^{2} x^{2} + p^{2} y^{2} + p^{2} z^{2} + q^{2} x^{2} + q^{2} y^{2} + q^{2} z^{2} + r^{2} x^{2} + r^{2} y^{2} + r^{2} z^{2}$
$2 p q x y + 2 q r y z + 2 p r x z = p^{2} y^{2} + p^{2} z^{2} + q^{2} x^{2} + q^{2} z^{2} + r^{2} x^{2} + r^{2} y^{2}$
How do i solve it?

Addyson Bright 2022-09-14

Find $| 2 a + 3 b + 2 \sqrt{3} (a \times b) |$ where a,b are perpendicular unit vectors.
My attempt:
$(| 2 a + 3 b + 2 \sqrt{(} 3) (a \times b) |)^{2}$
$(a . b = 0, | a | = | b | = 1), \hat{n}$ is normal vector to a and b.
$4 + 9 + 12 + 8 \sqrt{3} (\pm 1) + 12 \sqrt{3} (\pm 1)$ $25 \pm 8 \sqrt{3} \pm 12 \sqrt{3} |$ I have to eliminate one case $25 - 20 \sqrt{3}$ and then my answer should be $(25 + 20 (3)^{1 / 2})^{1 / 2}$ or $(25 + 4 (3)^{1 / 2})^{1 / 2}$ or $(25 - 4 (3)^{1 / 2})^{1 / 2}$
Is this right way to solve the problem. I am given only one place to anwer and I have found three answers.

Hagman7v 2022-09-14

I have the following representation:
$(a + β^{'} X)^{2}$
where a is a constant and both $β$ and X are k by 1 vectors.
The representation is further written as follows:
$a^{2} + 2 a β^{'} X + X^{'} β β^{'} X)$
While the first part is clear to me, I struggle to understand why it is shown once with ′ (X′ and $β^{'}$ ) and once without ′ (X and $β$ ).

hommequidort0h 2022-09-14

How we can generate a random vector $E = [e_{1}, e_{2}, e_{3} . \dots, e_{N}] \in R^{N}$ such that $\sum_{i}^{N} e_{i} = T$ and $0 \leq e_{i} \leq d_{i}$ $\forall i \in 1, 2, 3, \dots, N$ where di specifies the upper bound on each element of the vector. Is this problem over constrained?

Aidyn Crosby 2022-09-07

The angle between u,v is 120 degrees, with $| | u | |$ =4, $| | v | |$ =3
I need to calculate $| | (u - v) | |$ and $| | (u + v) | |$ . I used $(| | u | |^{2} + | | v | |^{2} - 2 | | u | | | | v | | \cos (x))^{2}$ but still not sure.

mriteyl 2022-09-07

If A,B,C,D be 4 points in a space and satisfy $| \vec{A B} | = 3, | \vec{B C} | = 7, | \vec{C D} | = 11, | \vec{D A} | = 9,$ Then value of $\vec{A C} \cdot \vec{B D} =$
what i try
Let position vector of $A (0), B (\vec{b}), C (\vec{c}), D (\vec{d})$
Then $\vec{A C} \cdot \vec{B D} = | \vec{A C} | | \vec{B D} | \cos θ$
where $θ$ ia an angle between $\vec{A C}$ and $\vec{B D}$
How do i solve it help me please

pokvarilaap 2022-09-07

Determine for all values of the parameter p whether the statement $d \in< a, b, c >$ holds for the vectors $a, b, c, d \in R^{4} :$ $a = 3, - 1, 2, 1)^{T}$ , $b = (15, 8, 8, 7)^{T}$ , $c = (12, 6, 7, p)^{T}$ , $d = (6, 8, - 9, 12)^{T}$
d is a subspace of a,b,c ? How can we find all values of the parameter p?

Sanai Ball 2022-09-07

I need to find the equation of a plane that passes through the point with position vector $A = 5 i + j + 3 k$ and the line $l_{1} = 2 i - 8 j - 17 k + λ (i - 3 j - 4 k)$
I found the normal by finding the cross product of $A * [(5 i + j + 3 k) - (2 i - 8 j - 17 k)]$ which turned out to be $n_{1} = - 24 i - 32 j + 18 k$ . Then I used $r \cdot n = a \cdot n \to (5 i + j + 3 k) \cdot (- 24 i - 32 j + 18 k) = - 98$
However, it turns out that the above value should be 49 rather than −98. I realize that it can be divided by -2 to get the answer; however, I don't understand why this happens.

daniko883y 2022-09-07

$| \vec{r} - {\vec{r}}^{'} |^{2} = r^{2} + r^{' 2} - 2 r \cdot r^{'} \cos (θ)$
How is this possible?
I could understand why, if we had brackets instead of abs values. But as it is I cannot understand how to operate with the square of an abs. value

aurelegena 2022-09-07

Is this true or false? If false, why is it false? An equation of the form $a x_{1} + b x_{2} + c x_{3} = d$ can be visualized as a line in $R^{3}$ . I think it's false because an equation in this form it should be a plane in $R^{3}$ not a line. Is that right?

tonan6e 2022-09-07

How to multiply out brackets when they contain vectors
I am keen to know the general rule for any vector:
$(x - y) (x - y)^{T}$
How does it relate to $(x - y) (x - y) = x^{2} - 2 x y + y^{2}$
Thanks for your help

Vrbljanovwu 2022-09-06

Let $α \in Λ^{p} L$ , which is p-th power of L, where L is linear space of dimension equal to n. Let us consider the following map $f_{α} : L \to Λ^{p + 1} L$ given with the formula $f_{α} (σ) = α \land σ$ , where $\land$ is a wedge product.
Prove that if $α, β \in Λ^{p} L$ and p $f_{α} = f_{β} ⟺ α = β .$
The part where we assume $α = β$ is easy, but what about another implication? Does anything come your minds? This was the first thing I thought about:
$α_{1} \land \dots \land α_{p} \land σ = β_{1} \land \dots \land β_{p} \land σ ⟺ (α_{1} \land \dots \land α_{p} - β_{1} \land \dots \land β_{p}) \land σ = 0.$
I am stuck here, but maybe I don't see something very obvious about this.

Janessa Benson 2022-09-06

I have two vectors:
$A : {51.031, - 102.062, 51.031} (| A | = 125)$
$B : {2, - 4, - 1} (| B | = \sqrt{21} \approx 4.58)$
I am trying to find the amount of $\vec{A}$ in the $\vec{B}$ direction, so I used dot product with the coordinate method:
$(x_{1} \times x_{2} + y_{1} \times y_{2} + z_{1} \times z_{2})$
$([2 \times 51.031] + [- 4 \times - 102.062] + [- 1 \times 51.031]) = 459.279$
If the dot product is supposed to find the magnitude of a vector that is pointing in the direction of another vector, how do I get a result that's more than 3 times the length of the longest vector?
What am I doing wrong?

Kwenze0l 2022-09-06

Let ${\vec{V}}_{1}$ and ${\vec{V}}_{2}$ are two vectors such that ${\vec{V}}_{1} = 2 (\sin α + \cos α) \hat{i} + \hat{j}$ and ${\vec{V}}_{2} = \sin β \hat{i} + \cos β \hat{j}$ , where $α$ and $β$ satisfy the relation $2 (\sin α + \cos α) \sin β = 3 - \cos β,$ Find Value of $(3 \tan^{2} α + 4 \tan^{2} β)$

drobtinicnu 2022-09-06

I am trying to prove the following expression:
By expanding the RHS of the expression, show that
$\frac{d \hat{T}}{d t} = \frac{r^{'} \times (r^{″} \times r^{'})}{{‖ r ‖}^{3}}$
where $\hat{T}$ is the unit tangent vector to a curve, i.e. $\hat{T} = \frac{r^{'}}{‖ r^{'} ‖}$
I used the vector triple product to find the numerator which gives me $r^{″} (r^{'} \cdot r^{'}) - r^{'} (r^{'} \cdot r^{″})$ I know that $r^{'} \cdot r^{'} = {‖ r^{'} ‖}^{2}$ , but I do not know how to simplify the second bracket.
Does anyone have any suggestions or could push me in the right direction?

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