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Homework Help: Finding the magnitude of a vector

  1. Jul 30, 2016 #1
    1. The problem statement, all variables and given/known data

    If possible, refer to attached image rather than whats written below (as the image is easier to understand).
    Triangle ABC is equilateral and AD = BE = CF.
    Let u, v and w be unit vectors in the directions
    of AB, BC and CA respectively.
    Let AB = mu and AD = nu.
    i) Find BC, BE, CA and CF.
    ii) Find |AE| and |FB| in terms of m and n.

    I don't understand why I can't use the Pythagoras equation to find their magnitudes. I've done hundreds of questions similar to this and have found the magnitudes of more vectors than I can remember, and this is the first time I've ever come across a vector where the equation doesn't apply.

    mag vec.PNG

    2. Relevant equations

    |vector| = |ai+bj+ck| = sqrt(a^2 + b^2 + c^2)

    3. The attempt at a solution
    I've attached a photo of my working. Basically, AE=AB+BE=mu+nv, so |AE|=sqrt((mu)^2 +(nv)^2)
    I know this is the wrong answer and wrong method. The correct method is: |AE| = sqrt((mu + nv)^2) = sqrt(m^2 -mn + n^2), as |u|=|v|=1
    But I don't understand why that's the correct method! My tutor tried explaining it to me, he said that u, v, w are unit vectors but aren't "orthogonal" (I have little knowledge of what that means, I tried looking it up but all I could understand is that it has something to do with the vectors being right angles to each other??) and thus I can't use the i-j-k definition of magnitude for these vectors. That's all he said and I still don't understand graphically/visually how that works. Could you please provide a picture of what vector I can use the pythagoras theorem on and what vector I can't use it on? I'm a visual learner. As I wait for your responses, I will draw an accurately scaled version of this problem on a large sheet of paper and try to manually figure out the magnitude and see why the pythagoras theorem doesn't work. By the way, I have very little math foundation, so please avoid using complex terminology.
    mag vec 2.jpg
  2. jcsd
  3. Jul 30, 2016 #2


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    Hello Matthew1994. Welcome to PF.

    In short: The Pythagorean Theorem only works for right triangles..

    a2 + b2 = c2 works when a and b are the sides that form a right angle, and c is the side opposite the right angle, a.k.a. the hypotenuse.
  4. Jul 30, 2016 #3
    I already know that. What im asking is how is that not valid in this case? If we were to draw the triangle onto a graph and plot the points relevant to the question on that graph, we would be able to use the pythagoras theorem to find the distance between two points even if they arent right angles to each other. Here we're given the locations of the points, but we cant use the pythagoras theorem, and i want to know why. Whats the difference betwen this equilateral triangle and an equilateral triangle that i draw on a graph? Because one can use the pythagoras theorem to find AE and the other cant use it..
  5. Jul 30, 2016 #4
    The Pythagorean Theorem only works for right triangles (2 sides are perpendicular). In this particular problem you are trying to insist that the unit vectors, u and v, are perpendicular to one another (like a right triangle) which is false (they have 60 degrees between them).
  6. Jul 30, 2016 #5
    I'm still confused because if A=(x1,y1) and E=(x2,y2) then the distance between A and E is sqrt((x2-x1)^2 +(y2-y1)^2) not sqrt(((x2-x1)+(y2-y1))^2)... whats the formula for the distance between 2 vectors which arent perpendicular? Nowhere in my textbooks nor on the internet does it say. It always says the pythagoras theorem.
  7. Jul 30, 2016 #6
    so if the unit vectors being used are not perpendicular you use the more generalized 'law of cosines formula'. You can find some info on it here https://en.wikipedia.org/wiki/Law_of_cosines. Have you been introduced to the dot product in your course yet? If you have I can give you a better understanding.
  8. Jul 30, 2016 #7
    Yes i know the dot product. A•B=|A|*|B|cos(angle°) but i have no idea what the purpose of the dot product is in a graphical visual sense
  9. Jul 30, 2016 #8
    So the dot product is very useful because it will give you a general formula for the magnitude of a vector. The magnitude of a vector is defined to be |A|=√(A⋅A). In the case of the unit vectors being perpendicular this simplifies to Pythagorean's Theorem. If they are not, this becomes the law of cosines.

    So for your specific problem you would use the law of cosines which is similar to |AE|=√(m^2+n^2) but with an extra term under the square root. Make sense?
  10. Jul 30, 2016 #9


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    This was in response to my answer to post #2
    Well, yes you can draw the triangle onto a grid and plot the points relevant to the question on that graph.

    If you draw ##\ \vec{AC}\ ## along the x-axis, then w = -i , the x unit vector. However, unit vectors u, and v are not perpendicular to each other nor to w .The x component of each of these is 1/2 .From that you can find the y-components.
  11. Jul 30, 2016 #10
    Haha i will just take that as truth and just memorize it. Ugh its so annoying how my textbook didnt mention this beforehand. Thanks guys! :)
  12. Jul 30, 2016 #11

    Ray Vickson

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    Which post are you replying to here? Please use the "quote" button, so that the different posts are identified and readers can navigate through the thread.
  13. Jul 30, 2016 #12
    Oh i meant everyone who participated here in general. I didn't really get the understanding that i hoped I'd get coming here and so I'm just thanking everyone and leaving.
  14. Jul 30, 2016 #13


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    When you say you plot points on a graph, I assume you mean using Cartesian coordinates (x and y). By definition, the x and y directions are at right angles to each other.
    Suppose you want the distance from A=(x1, y1) to B=(x2, y2). Consider the point C=(x1, y2). This is on the same horizontal line as A and the same vertical line as B. So you have a right angled triangle ABC. Pythagoras' theorem therefore applies and gives the distance AB with the formula you are familiar with.
    In short, your method of plotting on a graph and reading off coordinates creates a suitable right angled triangle.
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