Graphing a vector

Question

I have a vector (in physic) designated asF1=250cos(60)i+250cos(60)j+250cos(45)k, and i would like to see it in a 3D graphic with the axis centered at the origin, after what i would include other vector from there. But i have been unable to graph it. I would like it to look like this:

Vector in Space

(first time user, would appreciate as much info as possible)

Thanks in advance

Seb.

"Please help me,I've been able to do this ..." is highly preferred here to "Please, solve my problem". What have you tried? — Dr. belisarius, Feb 27 '15 at 17:37
Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the [faq]! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Dr. belisarius, Feb 27 '15 at 17:38
Ive tried Plot3D, Line, Vector...Plot3D[Line[{Cos[60], Cos[60], -Cos[45]}], {x, -20, 20}],Plot3D[(250 cos[60] + 250 cos[60] - 250 cos[45])*x, {x, -20, 20}], stuff like that but i know its not the good tipping, because i,j,k are not variable. but i can`t seem to find any good answer either here or on the Mathematica library, though it seem pretty simple and basic... — Sebastien Comtois, Feb 27 '15 at 17:50

score 0 · Accepted Answer · answered Feb 27 '15 at 17:59

0

Graphics3D[Arrow[{{0, 0, 0}, {Cos[60], Cos[60], -Cos[45]}}], 
 PlotRange -> {#, #, #} &@{-1, 1}, Axes -> True, Boxed -> False, 
 AxesOrigin -> {0, 0, 0}]

Mathematica graphics

answered Feb 27 '15 at 17:59

Dr. belisarius

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Thanks enormously, i will be able to catch up from here. You are really helpful. – Sebastien Comtois Feb 27 '15 at 18:20
May i ask why you used {#,#,#}&@{-1,1}...what does it do, what does it stand for ? – Sebastien Comtois Feb 27 '15 at 18:22
@SebastienComtois Copy it to your notebook and evaluate it – Dr. belisarius Feb 27 '15 at 18:29
There is something wrong, it`s not in the good quadrant... it is supposed to be in {+,+,-} not in {-,-,-}, how is it possible that a vector(arrow) of origin {0,0,0} and end point at { Cos[60], Cos[60], -Cos[45] } can end up in that orientation ? – Sebastien Comtois Feb 27 '15 at 18:39
and ive tried what you said but i dont understand what the &@ ...(and everything else related to it), stands for. – Sebastien Comtois Feb 27 '15 at 18:41
it say >>The specified setting for the option Graphics3DBoxOptions, PlotRange cannot be used. >> if i change the ### value, for the {-1,1}, if i understand correctly, i think it`s for box orientation...size...??? – Sebastien Comtois Feb 27 '15 at 18:42
@SebastienComtois Well, you need to spend some time with Mathematica. Try reading this Q&A http://mathematica.stackexchange.com/q/18393/193 – Dr. belisarius Feb 27 '15 at 18:42

Graphing a vector

1 Answers1