Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Missing Components and 3D Vector Resolution by System of Equations

  1. Sep 25, 2010 #1

    Hi Im writing a program to calculate 3D vectors for visualization and Im testing textbook problems such as:

    "If vector c forms an obtuse angle with the 0Y-axis and is orthogonal to both vectors

    a = 3i + 2j + 2k and
    b = 18i - 22j - 5k

    Find the components of c if its norm (magnitude or length) is 14

    2. Relevant equations

    None, just resolution of vectors by use of 3 x 3 system of equations

    3. The attempt at a solution

    We know c is perpendicular to a AND b, therefore by definition
    [1] the cross product of a x b is also perpendicular to a AND b

    So we can assume that they lie in the same vector path/direction.

    However they are not the same length / magnitutde so they are not the same vector

    Therefore we want a vector such that the magnitude of c = 14 where <c1, c2, c3> are components of c.

    Also by definition the dot product of two vectors will be zero if they are perpendicular to each other, so we can say:

    c * a = 0
    c * b = 0
    c * (a x b) = 196 | 14^2 is 196

    Since c and a x b lie in the same path but share different magnitudes they are not the same vector...we want one along the same path that HAS a magnitude of 14...so we can create a system of equations that will produce the values of c1 , c2, and c3 as scalar multiples of a x b such that the magnitude of of the equation would be 14

    Therefore we would have:

    3c1 + 2c2 + 2c3 = 0 |a = <3,2,2>
    18c1 - 22c2 -5c3 = 0 |b = <18,-22,-5>
    34c1 + 51c2 - 102c3 = 14^2 (or 196) |(a x b) = <34, 51, -102>

    After solving the system we would get the necessary values for <c1, c2, c3> for c.

    So, is this the correct way to approach such a problem or is this WAY too complex and\or unnecessary----or at worst INCORRECT?

    I really appreciate any insight into this as I'd like to know if this is the most efficient WAY to solve this problem.

  2. jcsd
  3. Sep 25, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    You don't need to solve three equations. Once you have axb=<34,51,-102> you know (as you said) "c and axb lie in the same path". That means c=t*<34,51,-102> for some constant t. Figure out what values of t will give c a norm of 14. You should find two, one positive and one negative. Now which one gives you a c that makes an obtuse angle with the y axis?
  4. Sep 25, 2010 #3
    Thanks for the reply!

    I guess its good to know that I am on the right track.

    I did think it was a little redundant to have 3 equations.

    So suffice to say, would I be able to use the n x m technique to solve a non-square matrix of three unknowns and one equation by way of least-norm solution?

    Now a question:
    If I solve the system this way (nxm matrices with least-norm)...would I still need to differentiate between positive and negative solutions?

    And the other method I am presuming you are referring to is to solve by letting
    |axb|^2 = 14^2 = |a|^2 |b|^2 - (a*b)^2
    = |a|^2 |b|^2 (1 - cos^2 (90 degrees))

    and getting the roots of the quadratic...

    ...But wouldnt this 2nd method just equal a zero since a and b are perpendicular and therefore a*b will be a zero answer?

    Thanks again and I appreciate the insight
  5. Sep 26, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    You don't need a matrix at all. You just have to solve |t*axb|=|t*<34,51,-102>|=14. Don't you? It's automatically perpendicular to a and b. You just have to get the right norm and the obtuse condition.
    Last edited: Sep 26, 2010
  6. Sep 26, 2010 #5
    Im kinda of inexperienced in solving systems or solving for values...I tried to solve for t in the method you mention...but I get these really weird values and they dont 'seem' right.
    I guess I am unsure of the way to solve for t (which makes me really feel inexperienced)

    For example I tried it this way:

    If the formula for magnitude for a vector <a1, a2, a3> is sqrt(a1^2 + a2^2 + a3^2)
    then for the case of <34,51,-102> it would be

    c= 14 = sqrt(34t^2 + 51t^2 + -102t^2) ...for some value of t

    therefore we can square both sides to get rid of the radical to work the equation getting

    196 = 34t^2 + 51t^2 + -102t^2 ....since t is a factor, we can add them all up getting

    196 = t^2(-14) dividing both sides by -14, I get

    -14 = t^2 ....then I 'make' a quadratic by moving over the t^2

    -t^2 -14 = 0 ....i then can complete the square to solve for the roots for t, by moving the negative factor out making -(t^2+14) = 0

    now I can complete the square to complete the quadratic... getting

    -1(t+7)^2 = 49 ....to get rid of the negative I can add and subtract '1' and then move the leftover to the other side getting:

    (t+7)^2 = 50 ....now I can solve and I get the 2 solutions you mention:

    t = 0.07107
    t = -14.07107

    since the vector c needs to be obtuse to the 0Y axis and have a norm of 14 we need the values of <34,51,-102> to be dramatically reduced to get a norm of 14 so we choose:

    t = 0.07107

    so t*<34,51,-102> = <2.41638, 3.62457, -7.24914>

    However my norm ONLY comes out to where

    |0.07107*<34,51,-102>| = 8.45733

    Any ideas where I went wrong?
  7. Sep 26, 2010 #6

    You need to review basic algebra, if it's not a distraction.
    Plus with vectors you need to imagine mentally vectors in the space.
    Imagining them, rotating them, stretching and shortening them.

    There's no need of.

    You found this values of t by solving the equation above.
    What about making a quick check to see if they are correct ?

    I find it hard to imagine how 0.07107^2 can give -14
    Do a lot of training from basics.
    Last edited: Sep 26, 2010
  8. Sep 26, 2010 #7
    Thanks for the reply, it still feels a little cryptic to me, since the formula for finding the magnitude of vectors is

    sqrt(a1^2 + a2^2 + a3^2)

    so wouldnt t* <34,51,-102> = <34t,51t,-102t> => sqrt(34t^2 + 51t^2 + -102t^2) ???

    //then the equation to solve would be

    14 = sqrt(34t^2 + 51t^2 + -102t^2) right?

    So for this part how would it be incorrect...as for the basic algebra what subset of algebra are you referring...do you think it may just be more straightforward to use the 'Least Norm Matrix' method to get the components I guess?
  9. Sep 26, 2010 #8
    sqrt(a1^2 + a2^2 + a3^2)

    Try answer...in your example
    What is "a1" ?

    Write the complete vector with t inside, and the base vectors inside.
    Using Latex helps a lot to be neat and sharp
  10. Sep 26, 2010 #9
    Wait a sec...I see what you are saying...your suggestion, Quinzio, by imagining it in my head worked...I found the value of t that makes the equation work...HOWEVER, the outlined method I posted produces some interesting results:

    For example,

    by finding

    t = 0.07107
    t = -14.07107
    and choosing t = 0.07107

    then multiplying |0.07107*<34,51,-102>| = 8.45733 equals the dividing factor 1/ 8.5 ( 8.45733 rounded) i need for <34,51,-102> to work such that

    | 1/ 8.5*<34,51,-102> | = 14

    I find that very weird and interesting??? Why is this so...can I use this as an algorithm?

    Also what is Dick referring to when he says
    How can you find/ or how does there exist two values for 't'???

    Thanks again for all the assitance...I also dont mind at all reviewing learning much of what I have skipped over so let me know specifics if you can pinpoint them
  11. Sep 26, 2010 #10


    User Avatar
    Science Advisor
    Homework Helper

    It's less weird and interesting than it is wrong. If you buy that the vector you are looking for has the form c=t*<34,51,-102> then you want to solve 14=|c|=|t|*|<34,51,-102>|. This a lot easier than you are making it out to be.
  12. Sep 27, 2010 #11
    I see what you are saying Dick and I do agree I was approaching this problem from all angles except the most obvious and easiest one

    ...however, what I was trying to do was find a general way of solving a problem of:
    decomposing the vector c iff a AND b were not orthogonal or if a and b componenets weren't even known
    (I am still trying to find an algorithm for this type of scenario---do you know any techniques) the example discussed here was easier because a AND b were orthogonal
  13. Sep 27, 2010 #12


    User Avatar
    Science Advisor
    Homework Helper

    It's already general. You don't need that a and b are orthogonal to know that axb is orthogonal to both a and b. And if c is also orthogonal to a and b, it must be a multiple of axb.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook