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Vectors Help

  1. May 21, 2012 #1
    i'm probably posting in the wrong section but idk where else to post this :/
    I need help with a few questions !

    1. The problem statement, all variables and given/known data

    Q1: if vector AB =[3,-5,7] and A(-4, 3, 0), state the coordinates of B

    Q2: State a set of vectors that spans [itex]R^{2}[/itex] other than i or j

    Q3: let a represent vector a, b vector b, and m be a constant. Given a=[1,2] and b= [2,5], determine m if 2a+mb= [8,19]

    Q4: Express vector u as an algebraic vector give the magnitude of vector u = 36 and it has a bearing of 120 degrees.

    Q5: state x & y give vector w=[x, 0], vector r=[4, y] and -3w-5r= [2, 1]

    Q6: determine the point on the y axis equidistant from the points (2, -1, 1) and (0, 1, 3)

    q7:solve for x given vector u=[3x, 7]; vector v= [5x, x]; the magnitude of vector u plus vector v = 10x


    2. Relevant equations



    3. The attempt at a solution

    Q1: is it (-7, 8 -7)?
    Q2: does [7,9] work?
    Q3: this one idk what to do...
    Q4: idk how to do this question either..
    Q5: i wrote my answer as x= 22/3 and y= -1/5 and got a half out of one... where did i go wrong?
    Q6: is started off finding the midpoints my doing (2 + 0)/2, (-1+1)/2, and (1+3)/2. but don't know what to do after..
    Q7: i started off by adding 3x and 5x which is 8x. then did x+7. then did the square root of (8x)^2 and (x+7)^2, to solve for x... but it didn't work out nicely...
     
  2. jcsd
  3. May 21, 2012 #2
    roger Roger, what's your vector Victor!

    I'll give you a hand here;
    1) no
    If AB is the vector pointing from A to B and A is the vector pointing from 0 to A, what happens when we add A and AB?

    2)[7,9] is only one vector, the space generated by this is just a line (aka [itex]R^1[/itex]

    3)This is just a set of two linear equations for one unknown
    if a = [1,2] and b = [2,5]
    2a+mb = [8,19] says that
    2*[1,2] + m*[2,5] = [8,19]
    Which is equivelant to the statement
    2*1 + 2m = 8
    2*2 + 5m = 19
    (Since addition of vectors is componentwise)

    4)Picture a triangle with hypotenuse of length 36 at a 120 degree angle from the x axis, how would you find the x and y components of the point at the end of the hypotenuse (what do Cos and Sin do?)

    5)This is the same thing as question 3

    6)Look at the problem componentwise, what is the midpoint of 2 and 0, what is the midpoint of 1 and -1, what is the midpoint of 1 and 3?

    7)Are you sure you copied this correctly, I didn't get a very nice answer for x either, [itex]\frac{1}{65} \left(-2+i \sqrt{3181}\right)[/itex]
     
  4. May 21, 2012 #3
    Q1: so the answer is, [-1, -2, 7]?

    Q2: so to get vectors in two space you need to sets? so... [7,9] and [13,2]?

    Q3: m= 3? :D

    Q4: do you just express vector u in terms of the sin or cos law?

    Q5: oooh k

    Q6: the mid points are 1, 0, 2, i got that but idk what to do with those numbers after... or is that the answer?

    q7: yea, nvm...
     
  5. May 21, 2012 #4
    1) Yes, I hope you understand WHY this is though, rather than just adding them and hoping for the best

    2) Yes, again, I hope you understand WHY this is. For a set of vectors to span a vector space, you need to be able to write ANY vector in the space as a combination of vectors in the set. With just one vector, say [7,9] you can only write vectors of the form [7a,9a]

    3)bingo!

    4)Yes

    6)Where does the point [1,0,2] lie?
     
  6. May 21, 2012 #5
    1 on the x axis, 0 on the y and 2 on the z

    so the answer is 0 ? :s
     
  7. May 21, 2012 #6
    you're looking for a point on the y axis
    1. What must the x and z components be if the point is to lie on the y axis
    2. How would you define a plane that is perpendicular to the line between the two points and passes the midpoint?
    3. Using the above, what do you think your answer should be?
     
  8. May 21, 2012 #7
    to lie on the y axis z and x have to be 0. the the coordinates should be [0, y, 0]?

    i don't get step 2 though... stuck....

    edit; nvm do you just do, (2-0)^2 + (-1 -y)^2 + (1-0)^2 = (0-0)^2 + (1-y)^2 + (3-0)^2

    and solve for y? and y should be 1?
     
    Last edited: May 21, 2012
  9. May 21, 2012 #8
    you could also do that, yes :biggrin:

    my idea was to find the plane of all equidistant points and then find the one that lies on the y axis
     
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