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Linear combinations

  1. Jun 8, 2006 #1
    struggling with these problems:

    1. determine if the following are:
    i) collinear: A(0, 3, 2), B(1, 5, 4) and C(3, 9, 8)
    ii) coplanar: A(1, 4, −5), B(2, 12, −8), C(4, 6, − 4) and D(5, 3, −2)

    i know that TWO vectors are collinear if it is possible to express one as a scalar multiple of the other. i think that with three vectors, it would be the same; that one vector could be expressed as a combination of the other two (i.e. A = kB + tC). if i found consistent answers for k and b, would that mean that all three are parallel?
    similarly for ii), im assuming that 4 points are coplanar if one can be exprssed as a combination of the other three (i.e. A = kB + sC + tD), but this would yield 3 equations with 3 unknowns which i do not know how to solve for...is this method wrong then?

    2. vectors u=(i+j+k) and v=(i−2j+k), find a vector w, of length sqrt(2), that is perpendicular to both u and v.

    im guessing i do cross product to find a vector perpendicular to both u and v and then find a vector parallel to the result of the cross product that has magnitude sqrt(2). how i would i express this in an equation to solve for the coordinates of w?
  2. jcsd
  3. Jun 9, 2006 #2


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    Get the determinant of the matrix formed by the vectors, if it's 0 then they are linear dependent (colinear or coplanar, depends on what are you looking for).

    Another way for coplanar is if the triple scalar product of the vectors is 0, then they are coplanar. And obviously for colinear the cross product of two vectors is 0.

    For the second problem, do the cross product and then calculate a unit vector in the direction of your resultant vector, then simply multiply the magnitude which each of the components of the unit vector.

    [tex] \vec{c} = \vec{u} \times \vec{v} [/tex]

    [tex] \vec{\lambda} = \frac{\vec{c}}{|\vec{c}|} [/tex]

    [tex] \vec{w} = w \vec{\lambda} [/tex]
    Last edited: Jun 9, 2006
  4. Jun 9, 2006 #3
    thanks for your help so far. however, we haven't learned how to do matrixes yet and since there are 4 vectors, i can't really take the triple scalar product of the vectors. for 1i), you said if taking the cross product of the two vectors is 0 then its collinear, but again i have 3 vectors. the number of vectors was my initial concern to begin with.
  5. Jun 9, 2006 #4


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    No, you're confused with lines I think (which are parallel if their directional vectors are multiples of eachother). Through 2 points, there is always a line! Using this, you could set up the line through 2 of the 3 points and check if the third is on it. A similar approach can be used for the plane, but this isn't necessarily the fastest way.

    You're starting of right: the cross product will result in a vector perpendicular to u and v. If you normalize this vector, it'll have length 1. Can you then 'scale' it to length sqrt(2)?
  6. Jun 9, 2006 #5


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    No, it does not! It simply means that they are coplanar (all lie in the same plane). However, I have another question: are A, B, and C vectors or points?? I suspect that they are points and that you need to determine whether the vectors AB and AC (or AB and BC or AC and BC) are parallel.
    Okay, here you specifically referred to "4 points"- so the question is not about A, B, C, and D but about AB, AC, and AD.

    You don't solve an equation! You do exactly what you just said: find the cross product of the two given vectors, then divide by its length to get a unit vector and multiply by sqrt(2) to get a vector of length sqrt(2).
    Last edited by a moderator: Jun 11, 2006
  7. Jun 10, 2006 #6
    ok, thank you guys for all your help, i fixed up both those quesitons.

    now however, i have a couple more questions:

    1. Given the vectors a=(0, 1, 5), b=(2, 1, −4) and c = (6, 4, 0):

    a) Prove that the three vectors form a basis in V3

    so i figured that if they are to form a basis in v3, they will be coplanar and therefore the vectors can be expressed in this fashion:
    a = kb +tc

    (0, 1, 5) = k(2, 1, −4) + t(6, 4, 0) which equates into three seperate equations:

    0 = 2k + 6t
    1 = k + 4t
    5 = -4k

    from the third eqn, we get k = -5/4 and subbing back into eqn's 1 and 2, i found t to be 5/12 and 9/16 respectively. obviously these are not the same and their basis proves invalid. however the way the question is stated implies that they are supposed to form a basis. is there a typo in the question or have i made a calculation error somewhere?

    b) Express the vector (11, 9, −1) as a linear combination of a, b and c

    would i formulate a general statement saying (11,9,-1) = ka + tb + sc? forcasting forward, i can see that i will have 3 equations and 3 unknowns which i don't necessarily know how to solve yet...what happens in this situation or if the method is completely wrong, how do you express the vector as a combination of the other three?
  8. Jun 10, 2006 #7


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    1) The fact that those 3 vectors should form a 3-dim basis means that they have to be linearly independent!

    2) The approach is fine, you have 3 equations in 3 unknowns.
    Last edited by a moderator: Jun 11, 2006
  9. Jun 10, 2006 #8


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    That's just completely backwards! You don't say what "V3" is but in order for any set of vectors to form a basis they must be independent, not dependent

    The fact that you cannot do this is precisely what proves this is a basis!

    Yes, (11, 9, -1)= k(0, 1, 5)+ t(2, 1, −4)+ s(6, 4, 0)= (2t+ 6s, k+t+4s,6k-4t) so 2t+ 6s= 11, k+ t+ 4s= 9, 6k-4t= -1. Solve those three equations for k, t, s. You can, for example, solve the first equation for s in terms of t: s= (11- 2s)/6, and solve the last for k in terms of t: k= (-1+ 4t)/s. Now replace k and s in k+ t+ 4s= 9 with those to get one equation for t.
  10. Jun 10, 2006 #9
    ok guys, even though i answered both the questions successfully now, i obviously don't have the greatest handle on the theory.

    so what does it mean to form a basis in 3d?

    also, if AB, AC and AD are linearly dependent, geometrically, what is the relationship between the points A,B,C and D. will they form a closed figure?

    sorry guys for all the questions, my lack of understanding has been partially due to taking this course through online correspondence which essentially means, i have to teach everything myself.
  11. Jun 10, 2006 #10


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    For an n-dimensional space V, you need exactly n lineairly independent vectors which span the space to have a basis, i.e. every vector v in V can be written as a lineair combination of the vectors of your basis, but the basis vectors itself have to be indepedent.

    Do you understand that?

    For the case n = 3, for example in R³, the standard basis is {(1,0,0),(0,1,0),(0,0,1)}.
    These are indepedent and every vector in R³ can be written as a lineair combination of these vectors, since an arbitrary vector (a,b,c) can be written as: a(1,0,0)+b(0,1,0)+c(0,0,1).

    Take another set: {(1,0,0),(0,1,0)}. This set isn't a basis for R³ because it only spans a twodimensional space. All vectors with a non-zero third component cannot be created. Although these two are independent, they don't span R³ -> no basis.

    Another one: {(1,0,0),(0,1,0),(0,0,1),(-2,3,4)}. This set spans R³ (just take the first three vectors and look at my example of the standard basis), but it isn't indepedent, since (-2,3,4) = -2(1,0,0)+3(0,1,0)+4(0,0,1) -> no basis.
  12. Jun 11, 2006 #11
    thanks TD, cleared some things up.

    ok new question. if you take a look at this parallelpiped

    http://img134.imageshack.us/img134/1029/parallelpied0rp.jpg [Broken]

    i'm supposed to mark the vectors p = a + 0.5b + c and r = a - b - c on it. i know how to geometrically add and subtract vectors but i dont know where i should choose my starting point. is there a conventional spot where i should start constructing p an r? or is it okay to start conveniently anywhere as long as i show where my resultant vector came from?

    Last edited by a moderator: May 2, 2017
  13. Jun 11, 2006 #12


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    The question isn't clear, unless an origin is given, from where you start of course!
    If you can choose, take the corner where c ends and a and b begin, for example.
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