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Concept question about magnitude of vectors

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    this is just a personal question that i have. i know that to find the magnitude of any vector with components, say, 3i + 4j, you need to add 3^2 + 4^2 and take the square root. i know this comes from the pythagorean theorem so, my question is, how is it possible to use the same method for finding the magnitude of a vector with three dimensions, i, j, and k. i know that it works with the pythagorean theorem... because i've done it, but it doesn't make sense to me why that would work.


    2. Relevant equations
    a^2 + b^2 = c^2


    3. The attempt at a solution
    see above
     
  2. jcsd
  3. Sep 16, 2009 #2
    Its quite simple, try for example to use a cube with side 4 and then calculate the distance from the upper left corner to the lower right corner on the opposite side of the cube.
     
  4. Sep 16, 2009 #3
    Imagine a two dimensional vector whose x component is the width of your computer screen measured from the bottom left to the bottom right corner and whose y component is the screen's height, measured from the bottom left to the top left corner. As you said, by the Pythagorean theorem, the length of a vector with those components is the square root of the sum of the squares of your screen's width and height. This vector points from the bottom left of your screen to the top right.

    Let's expand to three dimensions. Let the x and y component of your vector be the same. but now let the 3rd dimensional z component be the distance from your screen to the wall behind your monitor. This vector stretches from the bottom left corner of your computer screen to the top right corner of your computer screen to the wall behind that top right corner. Now we already found the length of the vector stretching from the bottom left to the top right: it is the square root of the width of your screen squared, plus the height of your screen squared.

    Imagine a straight (FIRST) line from the bottom left of your screen to the top right. Now imagine a (SECOND) line from the top right of your screen to the wall behind it. A line stretching from the START OF YOUR FIRST line to the END OF YOUR SECOND LINE completes a two dimensional triangle. The length of that stretching line is the square root of the sum of the squares of the legs.

    One of the legs is the length of your first vector (from the bottom left to top right) . This leg squared is the sum of the bottom left to bottom right vector squared. adding the two the square of the distance from your monitor to the back wall, then taking the square root gives this distance of the vector stretching from the bottom left front of your screen to the top right back corner. This is lengthy but I hope it helps.
     
  5. Sep 16, 2009 #4
    It actually does help, thanks so much to both of you for taking the time to explain this to me :)
     
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