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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
maybe
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i hope so
 one year ago

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

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

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

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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
yeah,
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
So 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.
 one year ago

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
2D vectors can represent the points of the cartesian plane. So the different vectors represent different points. And they fit this plane.
 one year ago

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

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

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

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

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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i suppose i think the different vectors look different, so they must be orthogonal somehow
 one year ago

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

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

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Scalar product of the vectors is just a particular case of the general multplication of the matrices.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
can you graph a matrix?
 one year ago

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

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
No. 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?
 one year ago

berlingots Group TitleBest ResponseYou've already chosen the best response.0
since 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.
 one year ago
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