Basic Electromagnetics Help: Vector Points & Triangle Medians Explained

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The discussion focuses on basic electromagnetics, specifically converting points into vectors and calculating triangle medians. A user seeks help with expressing points A(2,1,-5) and A(1,4,6), B(5,-3,0) as vectors, and finding the lengths of medians for triangle ABC with vertices A(2,1,3), B(3,-1,-2), and C(0,2,-1). The solution involves using the origin as a reference for position vectors and calculating vector lengths using the square root of the sum of squared coordinates. The responses clarify how to derive position vectors and compute lengths, emphasizing the importance of understanding vector representation in three-dimensional space. This foundational knowledge is crucial for tackling similar problems in electromagnetics.
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Hi, I'm Xairylle and I need a little help with this. I think it's kind of basic electromagnetics stuff but I don't know how this goes given that all I've got is a problem nothing else. I can't seem to get any of it and... it's not homework.

1. Express each of the given points into its equivalent vector
a) A(2,1,-5)
b) A(1,4,6) B(5,-3,0)

2. Find the lengths of the medians of the given triangles ABC
a) A(2,1,3) B(3,-1,-2) C(0,2,-1)

I need solutions and answers since I really can't understand. Please help if it's not too much trouble. Thank you.
 
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while this may not be homework, it is the sort of question which is very common in homework.
 
Each number in the bracket of three numbers is just representing a different dimension. E.g. (l, w, h)
 
If by "equivalent vector" you mean what I would call "position vector" it is the vector pointing from the origin, (0, 0, 0), to the given point, (x, y, z) and is written as xi+ yj+ zk.
 
First of all, you must know 2 points (usually given) in order to create the specific vector. If so, the vector's coordinates are given by the coordinates of the second point minus the coordinates of the first.
For example, find the coordinates of the vector AB where A(1,-2,4) and B(2,2,-1).
It's obvious that AB(2-1,2+2,-1-4) ---> AB(1,4,-5)

In your first exercise, the position vector is asked. The position vector can be created by only knowing one point. The other point is always the origin point O(0,0,0).
Here are your answers:
1)
a) OA(2-0,1-0,-5-0) ----> OA(2,1,-5)
b)AB(5-1,-3-4,0-6) ---->AB(4,-7,-6)

In your second exercise, the length of a vector is asked.
For example, if AB(2,-3,-1) you can find the length by:

squareroot[2^2 + (-3)^2 + (-1)^2]=squareroot(4 + 9 +1)= =squareroot(14)

Good luck https://www.physicsforums.com/images/smilies/smile.gif
:smile:
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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