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 one year ago
\[\newcommand\ve[1]{\vec{\boldsymbol #1}} % vector
\newcommand\uv[1]{\hat{\boldsymbol #1}} % unit vector
\begin{equation*}\ve A=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\langle a_1,a_2,a_3\rangle\\
\end{equation*}\]?
 one year ago
\[\newcommand\ve[1]{\vec{\boldsymbol #1}} % vector \newcommand\uv[1]{\hat{\boldsymbol #1}} % unit vector \begin{equation*}\ve A=\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\langle a_1,a_2,a_3\rangle\\ \end{equation*}\]?

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

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

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

klimenkov
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0can you explain why that is so ?

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

klimenkov
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0do the different vectors fit the same cartesian plane?

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

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

klimenkov
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/50f4096be4b0694eaccfaa5d

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

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

klimenkov
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0can you graph a matrix?

klimenkov
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0not even a 2x2 matrix?

klimenkov
 one year 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
 one year 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.
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