A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 3 years ago
\[\newcommand\ve[1]{\vec{\boldsymbol #1}} % vector
\newcommand\uv[1]{\hat{\boldsymbol #1}} % unit vector
\begin{equation*}\ve A=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\langle a_1,a_2,a_3\rangle\\
\end{equation*}\]?
UnkleRhaukus
 3 years ago
\[\newcommand\ve[1]{\vec{\boldsymbol #1}} % vector \newcommand\uv[1]{\hat{\boldsymbol #1}} % unit vector \begin{equation*}\ve A=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\langle a_1,a_2,a_3\rangle\\ \end{equation*}\]?

This Question is Closed

PeterPan
 3 years ago
Best ResponseYou've already chosen the best response.0If the column matrix is the same as the vector?

PeterPan
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand the question (if this is a question) :(

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0well i'm trying to understand the different notations,,

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2It depends on the situation. In one case you can use different notations for the vector and in another  you cant.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0can you explain why that is so ?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, I can. Do you know anything about matrix multiplication?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2So I show a case, when the row notation and the column notation should not be confused. Lets take two vectors: \(\vec{a}=\left(\begin{matrix}1\\2\end{matrix}\right)\) and \(\vec{b}=(3,4)\). And now multiply \(\vec{a}\) on \(\vec{b}\), and then \(\vec{b}\) on \(\vec{a}\): \(\vec{a}\vec{b}=\left(\begin{matrix}1\\2\end{matrix}\right)(3,4)=\left(\begin{matrix}3&4\\6&8\end{matrix}\right)\) \(\vec{b}\vec{a}=(3,4)\left(\begin{matrix}1\\2\end{matrix}\right)=3\cdot1+4\cdot2=11\) If we confuse the rows and the columns we will have the wrong result, because it is important to know where is the column and where is the row. But in other case, when we say about a vector in general, without multiplication, it is not so important to know if it is a row or a column.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0do the different vectors fit the same cartesian plane?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.22D vectors can represent the points of the cartesian plane. So the different vectors represent different points. And they fit this plane.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Do you have a concrete example, so I can help you?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0or should one of those be on the z axis?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2It is ok. Why do you think any of those must be on zaxis? One more question: how did you draw this pic?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/50f4096be4b0694eaccfaa5d

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Oh. I thought about it, but it is too long to draw pics in TikZ. Or you have good skills?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i just made my first 3d template on ti\(k\)z today

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i suppose i think the different vectors look different, so they must be orthogonal somehow

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0* i mean the notation is different

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2No. Different vectors on the plane are the vectors, that has different components. The notation doesn't play any role.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0ok so then, the operation of multiplying the vectors somehow chooses that the first vector to be a row vector for dot product, right?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Scalar product of the vectors is just a particular case of the general multplication of the matrices.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0can you graph a matrix?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2You have to know what a matrix interprete. It is a table of numbers. My answer is No.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0not even a 2x2 matrix?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2No. How do you think, what does this matrix will show on the plane? A components of the vectors can be read as the point, but what is a matrix?

berlingots
 3 years ago
Best ResponseYou've already chosen the best response.0since you written an arrow on top of the A, I assume it is a vector. So the vector is 3 dimensional in R^3. Let the a's equal x, y, and z standing for its components. The second is written in column form and the last one is written in row form. I hope this answers your question. They are both equal but written differently notation wise.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.