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Homework Help: Finding the angel between two vectors

  1. Apr 9, 2007 #1

    ssb

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    I have two vectors

    Vector A and B
    Axi = -3
    Ayj = 5.4

    Bxi = 1.3
    Byj = -3.9

    I want the angle between them so I do the dot product

    (-3)(1.3)+(5.4)(-3.9) = -24.96

    Therefore the angle between them is

    -24.96 = |-24.96|cos(theta)

    Arccos(-24.96/24.96) = theta

    Arccos(-1) = theta

    Theta = pi = 180 degrees.

    Is this correct? :confused:
     
  2. jcsd
  3. Apr 9, 2007 #2

    cristo

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    Your calculation for the dot product is correct, however you have used the wrong formula: the formula you need is [tex]\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta[/tex]. How do you find the magnitude of a vector? (i.e. [itex]|\vec{a}|[/itex])
     
    Last edited: Apr 9, 2007
  4. Apr 9, 2007 #3

    ssb

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    Ok so the magnitude of vector a would be
    = sqrt(5.4^2 + 3^2)
    = (3*sqrt(106))/5
    = 6.18

    Magnitude of B would be
    = sqrt(-3.9^2 + 1.3^2)
    = (13*sqrt(10))/10
    = 4.11

    therefore

    -24.96 = abs(4.11) * abs(6.18) cos(theta)

    Theta = arccos [(-24.96)/(abs(4.11) * abs(6.18))]

    theta = 90 degrees = pi/2 radians

    Does this look accurate?
     
  5. Apr 9, 2007 #4
    It can't be 90. That would imply the dot product is zero (see the formula mentioned)
     
  6. Apr 9, 2007 #5

    ssb

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    [tex]\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta[/tex]

    I guess im still confused on the [tex]|\vec{a}||\vec{b}|[/tex] part

    this would be finding the magnitudes then multiplying their absolute values together right? The magnitudes are found with pathegoran's theorem correct?
     
  7. Apr 9, 2007 #6

    cristo

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    What is the value of [(-24.96)/(abs(4.11) * abs(6.18))]?
     
  8. Apr 9, 2007 #7

    ssb

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    its just about -1.

    Find the angle between
    A = -3i + 5.4j
    B = 1.3i - 3.9j

    (-3 * 1.3)+(5.4 * -3.9) = -24.96

    -24.96 = ab cos(theta)

    therefore

    theta = arccos (-24.96/sqrt[(2^2+1.3^2)(5.4^2+(-3.9)^2)]

    theta = domain error when I try to calculate it.
     
  9. Apr 10, 2007 #8

    HallsofIvy

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    Where did you get that "2" in the square root? You should have
    sqrt[((-3.9)^2+ 1.3^3)(5.4^2+ (-3.9)^2)].
     
  10. Apr 10, 2007 #9

    ssb

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    When you replaced my 2 with a -3.9, did you mean to replace it with a -3?

    If this is the case then my questions have been answered!!

    thanks everybody for helping me with this most basic topic
     
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