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Dot product questions

  1. Mar 28, 2005 #1
    I have some questions from my geometry and discrete class..

    #1 Given a and b unit vectors,
    a) if the angle between them in 60 degrees, calculate (6a+b)●(a-2b)
    b) if |a+b|= root(3), determine (2a-5b)●(b+3a)

    #2 The vectors a=3i-4i-k and b=2i +3j-6k are the daignols of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.

    ** for number 2) a and b are vectors ... i,j,k are unit vectors
     
  2. jcsd
  3. Mar 28, 2005 #2
    Okay... did you make any attempt at the problems?
     
  4. Mar 28, 2005 #3
    1a)[tex] X dot Y = |X||Y|cos(\theta) [/tex]


    #2. Dot the two vectors together to make sure they are perpendicular. You should get 0. Subtracting the two vectors should give you a third vector representing the side.

    Then to find the angles, use the relationship:

    X dot Y = [tex]X_iY_i + X_jY_j + X_kY_k [/tex]

    and X dot Y = [tex]|X||Y|cos(\theta)[/tex]

    Where Y is your third vector, and X is one of your original vectors.
     
  5. Mar 28, 2005 #4
    how do you create those formulas .. do you copy them from word?
     
  6. Mar 28, 2005 #5
    Use the forum's latex capabilities. The tags are [ tex ] [/tex]
     
  7. Mar 28, 2005 #6
  8. Mar 28, 2005 #7
    heres what i tried

    1a) [tex] (6\hat{a}+\hat{b})\bullet(\hat{a}-2\hat{b}) [/tex]
    [tex] = 6|\hat{a}|^2 - 12\hat{a}\bullet\hat{b} + \hat{b}\bullet\hat{a} - 2|\hat{b}|^2 [/tex]
    [tex]= 6|\hat{a}|^2 - 11\hat{a}\bullet\hat{b}- 2|\hat{b}|^2 [/tex]
    [tex]= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|cos\Theta- 2|\hat{b}|^2 [/tex]
    [tex]= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|cos60- 2|\hat{b}|^2 [/tex]
    [tex]= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|(1/2)- 2|\hat{b}|^2 [/tex]


    not sure what to do next

    #2) did the dot product... gave me zero
    did a-c .. got (1,-7,5) .. did [tex] \sqrt{1^2+-7^2+5^2} = \sqrt[5]{3} [/tex]
    the back of the book .. has [tex] \frac{\sqrt[5]{3}}{2} [/tex] ... why is it over 2?
    then for one of the angles i did this:
    [tex] cos\Theta = \frac{\vec{a}\bullet\vec{c}}{|\vec{a}||\vec{c}|} [/tex]
    [tex] cos\Theta = \frac{26}{\sqrt{1950}} [/tex]
    [tex] \Theta = 54 [/tex]
    which is wrong
     
    Last edited: Mar 28, 2005
  9. Mar 28, 2005 #8

    Galileo

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    Okay, you'd know the answer if you knew what |a| and |b| are right?
    So can you find out what they are from the given data?
     
  10. Mar 28, 2005 #9
    im assuming your talking about |a + b| = root 3 ... which is only used for b) .. not a) .. so you DONT know what |a| and |b| are
     
  11. Mar 28, 2005 #10

    Galileo

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    No, I'm talking about question a). What can you say about the magnitudes of [itex]\vec a[/itex] and [itex]\vec b[/itex]?
     
  12. Mar 28, 2005 #11
    theyre 1? .. then youll end up getting 11/2 = 4 .. which ends up giving you 3/2 right?
     
  13. Mar 28, 2005 #12
    Yeah, |a| = |b| = 1 cos you're told they're unit vectors. 11/2 isn't 4 though :confused:.
     
  14. Mar 28, 2005 #13
    Now since it is given that a and b are unit vectors , therefore |a| = 1 and |b|=1

    so (6a + b).(a-2b) = 4 - (11/2) = -3/2

    Now are you able to get the result ?
     
    Last edited: Mar 28, 2005
  15. Mar 28, 2005 #14
    6i+j dot i-2j theta = 60

    mag a = [tex] \sqrt{6^2+1^2} = sqrt{37} [/tex]

    mag b = [tex] \sqrt{1^2+2^2} = sqrt{5} [/tex]

    a dot b = [tex] |a||b|cos(\theta) = sqrt{185}(1/2)[/tex]

    Sorry for the different variables.
     
  16. Mar 28, 2005 #15

    dextercioby

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    If u're using Latex to edit your formulas,then use the function "\frac{...}{...}" for writing fractions...

    U may wanna pay more attention with paranthesizing when necessary (v.post #7,the square roo t with 2 consecutive signs & no round brackets).

    Daniel.
     
  17. Mar 28, 2005 #16
    For #2, remember that vectors A and B and the adjoining vector C represent one quarter of the rhombus. A diagonal stretches across half of a rhombus, before you subtract the two vectors and dot them, you need to half them.

    a dot c

    [tex] \hat a = 3/2\hat i - 2\hat j - 1/2\hat k [/tex] with Magnitude: [tex] \sqrt{3/2^2+(-2)^2+(-1/2)^2} = \sqrt{9/4+4+1/4} = \sqrt{13/2} [/tex]

    [tex] \hat c = 1/2\hat i - 7/2\hat j + 5/2\hat k[/tex] with Magnitude [tex]\sqrt{1^2+(-7)^2+5^2} = 5\sqrt3[/tex]

    [tex] a \bullet c = ((3)(1) + (-4)(-7) + (-1)(5))(1/2) = (1/2)(3 + 28 -5) = 13[/tex]

    [tex] cos(\theta) = \frac{a\bullet c}{|A||B|} = \frac{13}{5\sqrt{39/2}} = 0.59 [/tex]
     
  18. Mar 28, 2005 #17
    The answer in book is right. The length of side1, [tex]|\vec{X}|[/tex] is obtained by

    [tex] \vec{X} = \frac {\vec{a} - \vec{b}}{2} [/tex]

    The length of side 2 is, [tex]|\vec{Y}|[/tex]

    [tex] \vec{Y} = \frac {\vec{a} + \vec{b}}{2} [/tex]

    The angle, [tex] \theta [/tex] between two sides can be found by X.Y and the other angle would be = 180 - [tex] \theta [/tex]
     
    Last edited: Mar 28, 2005
  19. Mar 28, 2005 #18
    90 - [tex] \theta [/tex] since the triangle is right.
     
  20. Mar 28, 2005 #19
    We are finding the angles between the two sides or say the angles that makes the two corners so it should be 180 - [tex] \theta[/tex]
     
  21. Mar 28, 2005 #20
    i just put the wrong type of slash before the sqrt, i'll edit it, but i think everyone understood what i did.
     
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