Learn Vectors with Examples, Equations, and Practice Problems

Show that if $u, v \in R^{m}$ , then $| u \cdot v | \leq | | u | |_{\infty} | | v | |_{1}$
By Cauchy-Schwarz,
$| u \cdot v | \leq | | u | |_{2} | | v | |_{2}$
Note also that $| | u | |_{1} \leq \sqrt{m} | | u | |_{2} \leq m | | u | |_{\infty}$ , and $| | v | |_{\infty} \leq | | v | |_{2} \leq | | v | |_{1}$ , so we get
$| u \cdot v | \leq \sqrt{m} | | u | |_{\infty} | | v | |_{1}$
How to get rid of $\sqrt{m}$ to get the desired inequality?

Damon Campos 2022-08-11

I am trying to prove that, given two linearly independent vectors $v_{1}, v_{2}$ , the sum of vectors $v_{3} + v_{4}$ (both being some linear combination of $v_{1}, v_{2}$ ) is linearly dependent with $v_{1}, v_{2}$
It makes sense to me intuitively since if we have $x_{1} v_{1} + x_{2} v_{2} = v_{3}$ and $y_{1} v_{1} + y_{2} v_{2} = v_{4}$ , then $v_{3} + v_{4} = x_{1} v_{1} + x_{2} v_{2} + y_{1} v_{1} + y_{2} v_{2}$ , but I'm struggling to formulate a proof using the definition of linear dependency. Any suggestions on how I might do so?

Makayla Eaton 2022-08-11

Imagine I have an arbitrary vector $v \in R^{n}$ and say it can be represented as
$v = (v_{1}, . . ., v_{n})^{T}$
where it's understood that the indexes are in increasing order. Imagine now the vector $u \in R^{n - 1}$ obtained by simply removing entry with index i of v. What's the best (and most "economic") way to represent such vector? My idea was to simply interpret the vector as a sequence and write
$u = {v_{j}}_{j \neq i}^{T}$
but I'm not sure whether this is correct and/or there is a better way of writing it.

Jenny Stafford 2022-08-10

Find the intersection of
$2 x_{1} + x_{2} + x_{3} - 3 = 0$ , $2 x_{1} + x_{2} + 4 x_{3} - 6 = 0$ , $2 x_{1} + x_{2} - 2 = 0$
So I calculated the determinant of the three planes and it's equal to zero, then I calculated the cross product between the two first planes and it gave me $3 x_{1} - 6 x_{2}$
The thorem in the course book says that if the determinant of the three planes is equal to zero, and cross product it's not zero, then there's two posibilities if $r_{4} - u p_{4} - v q_{4} = 0$ then the three planes intersect in a line l and $l = {(x_{1}, x_{2}, x_{3}) : S X = u p \times q}$ and S is a point in the line.
What is $r_{4}$ , $u p_{4}$ and $v q_{4}$ ?

imire37 2022-08-10

How to find a vector with two 3d points?
A:(2,3,1); B:(8,-2,4); C:(3,0,5).
find a so $\vec{A} \vec{B}$ is orthogonal to $\vec{A} \vec{B} + a . \vec{A} \vec{C}$
I think the answer is $(\vec{A} \vec{B}) . (\vec{A} \vec{B} + a . \vec{A} \vec{C}) = 0$ and then find a.(if this isn't the answer please tell me.)
but how can I write or find $\vec{A} \vec{B}$ . I think with matrices because that's what we are studying now.

Leia Hood 2022-08-10

Let $a, b, c, d \in R^{3}$ and $d i m (s p a n (a - d, b - d, c - d)) \leq 2$ (i.e. the vectors $a - d$ , $b - d$ , $c - d$ are linearly dependant). Prove that $a, b, c, d$ lie on an affine plane.

Makayla Eaton 2022-08-10

The derivation the following equation is given as a statement of D'Alambert's principle:
${\vec{F}}_{net} \cdot δ \vec{r} = m \frac{d^{2} \vec{r}}{d t^{2}} \cdot δ \vec{r}$
After which the following equation is used: $(x y)^{'} = x^{'} y + x y^{'}$ to rewrite the right hand side so that:
$m \frac{d^{2} \vec{r}}{d t^{2}} \cdot δ \vec{r} = m [\frac{d}{d t} (\frac{d \vec{r}}{d t} \cdot δ \vec{r}) - \frac{d \vec{r}}{d t} \cdot \frac{d δ \vec{r}}{d t}]$
What is x, what is y and how is the equation applied?

musicintimeln 2022-08-10

Prove that $\vec{p} \cdot \vec{q} = | \vec{p} | | \vec{q} | \cos a$ the angle between vector p and q.
I tried using law of cosines but I'm not supposed to do that since I need to prove law of cosines in the next exercise, also I think law of cosines is a consequence of this statement.

June Mejia 2022-08-10

Given that the vectors $v_{1}, v_{2}, v_{3}, v_{4}$ are linearly independent.
Are the following $b_{1}, b_{2}, b_{3}$ vectors linearly independent too?
$b_{1} = 3 v_{1} + 2 v_{2} + v_{3} + v_{4}$
$b_{2} = 2 v_{1} + 5 v_{2} + 3 v_{3} + 2 v_{4}$
$b_{3} = 3 v_{1} + 4 v_{2} + 2 v_{3} + 3 v_{4}$
What I've done so far
$k_{1} (3 v_{1} + 2 v_{2} + v_{3} + v_{4}) + k_{2} (2 v_{1} + 5 v_{2} + 3 v_{3} + 2 v_{4}) + k_{3} (3 v_{1} + 4 v_{2} + 2 v_{3} + 3 v_{4}) = 0$
then I don't know what to do. I don't even know whether the 0 above is supposed to be a vector or a scalar.

sapetih1i 2022-08-10

Can eigenvectors be scaled and still be eigenvectors?
Given the matrix:
$A = [\begin{matrix} 2 & 0 \\ 1 & 4 \end{matrix}] .$
The eigenvalues are:
$λ_{1} = 2,$
$λ_{2} = 4.$
To find the eigenvectors:
$v_{1} = nullity (A - λ_{1} I),$
$v_{2} = nullity (A - λ_{2} I) .$
Which yielded the following eigenvectors for $v_{1}$ :
$v_{1} = [\begin{matrix} - \frac{2}{3} \\ 1 \end{matrix}] .$
However, I have seen this alternative eigenvector for $v_{1}$ :
$v_{1} = [\begin{matrix} - 2 \\ 3 \end{matrix}] .$
Question: Are both eigenvectors for $v_{1}$ correct? In other words, can you scale an eigenvector by any real number and it still be an eigenvector of that matrix?

traquealwm 2022-08-10

If $u = (- 2, 0, \sqrt{2})$ and $v = (0, - 4, - \sqrt{2})$ , then $u \times v = (2 \sqrt{2}, - \sqrt{2}, 8)$
I got $(4 \sqrt{2}, 2 \sqrt{2}, 0)$ for the cross product but the answer is saying I'm wrong and I don't know why

cortejosni 2022-08-10

Find the normal vector N to $r (t) = ⟨ t, \cos t ⟩$ at $t = \frac{9 π}{4} .$
How do I find this normal vector? So basically I did what the feedback said. I found the derivative of each function in the vector and got.
$(1, - \sin (t))$
Then I got the magnitude:
$\sqrt{1 + \sin^{2} (\frac{9 π}{4})}$
Then I divided everything in my vector by that magnitude while putting in
$\frac{9 π}{4}$
like so:
$\frac{- \sin (\frac{9 π}{4})}{\sqrt{1 + \sin^{2} (\frac{9 π}{4})}}$
What my problem? Here's my answer:
$⟨ \sqrt{\frac{2}{3}}, - \frac{\sqrt{3}}{3} ⟩$

empalhaviyt 2022-08-10

How to find shortest distance between a line and y axis
The equation of line is $r = (3, 0, 0) + λ (3, 0, - 4)$
And the equation of the y axis is x=0 ,z=0
For y axis $r = (0, k, 0) + λ (0, 1, 0)$
I know Formula for calculating the shortest distance is : $| \frac{(a_{2} - a_{1}) . (b_{1} \times b_{2})}{(b_{1} \times b_{2})} |$
How do i calculate k so that i can find the shortest distance?

Aleseelomnl 2022-08-10

Find the plane through $(1, 2, 3)$ and parallel to the plane determined by the points $(2, 1, - 1); (1, 1, 1)$ and $(3, 3, 0)$

pominjaneh6 2022-08-09

Say you have ${\vec{v}}_{1} = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$ and ${\vec{v}}_{3} =$ anything.
Is the set ${{\vec{v}}_{1}, {\vec{v}}_{2}, {\vec{v}}_{3}}$ always considered linearly dependent because ${{\vec{v}}_{1}, {\vec{v}}_{2}}$ is?

Bierlehre59 2022-08-09

let $M \in M_{n} (R)$ be a matrix such that $M^{2} - M = 0$ , Then find $M^{- 1}$ (M inverse)?
My approach is:- Let M be an invertible matrix, $\Rightarrow$ $M x M^{- 1} = I_{n}$ , where $I_{n}$ is identity matrix of order n. So, $M^{2} - M = 0, \Rightarrow M^{2} = M, \Rightarrow M$ is an Idempotent Matrix. Now, multiple both sides with $M^{- 1} \Rightarrow M^{2} \times M^{- 1} = M \times M^{- 1} \Rightarrow M = I_{n}, \Rightarrow M^{2} = I_{n}$
But I am unable to proceed further to calculate the matrix. I don't know what steps I have follow to calculate $M^{- 1}$ of M.

crazygbyo 2022-08-09

Say I have the equation
$x = \vec{a} \cdot (\vec{b} - \vec{c})$
where $\vec{a}$ is a column vector, $\vec{b}$ , and $\vec{c}$ are row vectors, x is a scalar, and $\vec{a} \cdot (\vec{b} - \vec{c})$ is taking the dot product between $\vec{a}$ and $(\vec{b} - \vec{c})$ . Is it possible to solve for $\vec{b}$ ?
I would naively think I could do this as
$x = \vec{a} \cdot (\vec{b} - \vec{c})$
$x {\vec{a}}^{T} = | | \vec{a} | |^{2} (\vec{b} - \vec{c})$
$x {\vec{a}}^{T} = | | \vec{a} | |^{2} \vec{b} - | | \vec{a} | |^{2} \vec{c}$
$\vec{b} = \frac{x {\vec{a}}^{T} + | | \vec{a} | |^{2} \vec{c}}{| | \vec{a} | |^{2}}$
But this does not seem correct when putting in concrete values. What am I doing wrong?

Sydney Stein 2022-08-09

Let $[a \times b b \times c c \times a] = k [a b c]^{2}$ . Find k. Here, $[u v w] = u \cdot (v \times w)$
My attempt:
Writing scalar triple product as : $(a \times b) \cdot ((b \times c) \times (c \times a))$ . Not able to proceed next.

popljuvao69 2022-08-09

I have two points in cartesian coordinates $P_{1} = (1, 0, 2)$ and $P_{2} = (0, 2, 1)$ and need the distance vector from $P_{1}$ to point $P_{2}$ . The vector from origin to $P_{1}$ is defined as R while the vector from origin to point $P_{2}$ is defined as $R'$ . The vector I am looking for is then $R-R'$ between the first and the second one in space. I have forgotten linear algebra a bit so I am not sure of the proper way of doing it. Could you please help me out?
Is this right?
$R = (1, 0, 2) R' = (0, 2, 1) R-R' = (1, 0, 2) - (0, 2, 1) = (1, - 2, 1)$

lollaupligey9 2022-08-09

I know that $| A \times B | = | | A | | | | B | | \sin (θ)$ , wherein the case of 3D Vectors and I wanted to calculate $\sin (θ)$ from the previous formula I will get a [1x3] vector as a result, my question is, can the 3 values in the resulted sin vector be treated as 3 separate angles with the primary axes, say X,Y,Z?

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