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Homework Help: Vector Operations

  1. Jun 6, 2006 #1
    Q: Given the points A(1,6,-2), B(2,5,3) and C(5,3,2). Use two different vector methods to determine whether ΔABC is a right angle triangle.

    hmmm vector methods...i'm not quite sure what to do but i have some ideas. for a first method, could i take the magnitude of each side and see if the trig calculations hold valid (pretty weak method considering theres many assumptions here). i'm thinkign a better method would be dot and cross product. i.e. for dot product, see if any two sides produces a product of zero?

    i would just someone to guide me the right way so no time is wasted guessing around.

    thanks in advance!
  2. jcsd
  3. Jun 6, 2006 #2
    Hint: Use dot product for 1 method and cross product for the second, longer, method . Start by forming vectors AB, BC and CA .
  4. Jun 6, 2006 #3


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    Trig calculations? Do you mean sine and cosine? Since you don't know any angles I don't see how you could do that. Of course the Pythagorean theorem might help.

    Yes, that would work as a second method.

    Time is never wasted while "guessing around". The more you do it the more you learn.
  5. Jun 6, 2006 #4
    for dot product: how do i know which two vectors could potentially have an angle of 90 between them? would i have to try all 3 possibilities?

    for cross product: what am i doing here?

    HallsofIvy: what i meant for trig calculations was in fact pythag. to see if any two sides produced a third side that was the sum of the squares of the other two (i.e. a^2 + b^2 = c^2)

    you're also right about guessing around, that's how we learn i guess.
  6. Jun 7, 2006 #5


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    You have basically been given three vector equations in the question;

    [tex]\vec{OA} = 1i + 6j -2k[/tex]

    [tex]\vec{OB} = 2i + 5j -3k[/tex]

    [tex]\vec{OC} = 1i + 6j -2k[/tex]

    You now need to obtain three more vector equations; [itex]\vec{AB}[/itex], [itex]\vec{BC}[/itex], [itex]\vec{CA}[/itex]. Can you do that?

    For the dot product basically, your just going to have to try all three combinations (there are only three).

    More information on the cross product can be found at
    http://en.wikipedia.org/wiki/Cross_product" [Broken]
    http://mathworld.wolfram.com/CrossProduct.html" [Broken]
    Last edited by a moderator: May 2, 2017
  7. Jun 7, 2006 #6
    You can also try Pythagoras theorem using the magnitudes of the vectors as Hallsofivy explained .
  8. Jun 7, 2006 #7


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    You are given A(1,6,-2), B(2,5,3) and C(5,3,2). So [itex]\vec{AB}= (2-1)\vec{i}+ (5-6)\vec{j}+ (-2-3)\vec{k}[/itex], [itex]\vec{AC}= (1-5)\vec{i}+ (6-3)\vec{j}+ (-2-2)\vec{k}[/itex], and [itex]\vec{BC}= (5-2)\vec{i}+ (3-5)\vec{j}+ (2-3)\vec{i}[/itex]. Is the dot product of any two of those 0? The cross product isn't applicable here.
  9. Jun 7, 2006 #8
    Why not ? If triangle ABC is rt angled at B , then ,
    [tex]\vert(\vec{AB}\times\vec{BC})\vert = \vert\vec{AB}\vert \vert\vec{BC}\vert[/tex].
    This is surely a necessary and sufficient condition although a bit long .
  10. Jun 7, 2006 #9


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    I believe the cross product gives the vector which is perpendicular to both A and B.
  11. Jun 7, 2006 #10
    I was only taking the magnitude (modulus). Also sin(B) = 1 .
  12. Jun 8, 2006 #11
    yeah the cross product would work here...none of dot products produced a value of 0 so i guess its not right-angled. didn't make an arithmetic error did i?


    wait if i used the cross product here, then a x b = |a||b| for any vector combination as arunbg stated. but then the end result would be (x,y,z) = a number; how would i rationalize the equality there?
    Last edited: Jun 8, 2006
  13. Jun 9, 2006 #12
  14. Jun 10, 2006 #13
    yeah i noticed HallsOfIvy's mistake when reading it, i figured it was a typo. but even with the corrected values, i can't get it to work.

    these are the values im getting:
    AB • BC
    = 10

    AB • CA
    = -27

    BC • CA
    = -14

    i think it has something to do with the vectors im forming; i formed AB, BC and CA but HallsofIvy formed AB, BC and AC. im getting confused because dot product is taken from tip to tip anyways.
  15. Jun 10, 2006 #14


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    Are you sure you are calculating the dot product correctly? Here's my working for the first one;

    [tex]\vec{AB}= (2-1)\vec{i}+ (5-6)\vec{j}+ (2+3)\vec{k}[/tex]

    [tex]\vec{BC}= (5-2)\vec{i}+ (3-5)\vec{j}+ (2-3)\vec{k}[/tex]

    [tex]\vec{AB}{\mathbf\centerdot}\vec{BC} = 1\times 3 + -1 \times -2 + 5\times -1[/tex]

    [tex]\boxed{\vec{AB}{\mathbf\centerdot}\vec{BC} = 0}[/tex]

    Now what does this say about the vectors [itex]\vec{AB}[/itex] and [itex]\vec{BC}[/itex]? If you can't see it straight away, note that the cosine of the angle between them is given by;

    [tex]\cos\theta = \frac{\vec{AB}{\mathbf\centerdot}\vec{BC}}{|\vec{AB}||\vec{BC}|}[/tex]
    Last edited: Jun 10, 2006
  16. Jun 10, 2006 #15
    sorry i'm so careless (did 3+2+5 instead of 3+2-5). thanks hootenay (note that you have a mistake in your dot product, the first value should be 1 x 3 but i know it was a typo since your answer is still 0).
  17. Jun 10, 2006 #16


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    Ahh yes, careless error, I seem to be prone to them recently. Thank you for the correction. Now, have you thought anymore about my question;
  18. Jun 10, 2006 #17
    it says that the angle between the vectors AB and BC is 90 and hence a right-angle. is this what you were implying?
  19. Jun 10, 2006 #18


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    That's it. So now you've proved it with one vector method...:smile:
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