1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Finding the angel between two vectors

  1. Apr 9, 2007 #1

    ssb

    User Avatar

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar


    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

    User Avatar

    [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

    User Avatar
    Staff Emeritus
    Science Advisor

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

    ssb

    User Avatar

    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

    User Avatar
    Science Advisor

    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

    User Avatar

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...
Similar Threads for Finding angel between Date
Find the angle between 2 vectors Monday at 7:32 AM
Find the smallest value for the polynomial Apr 19, 2018
Find the value of the trigonometric sum Apr 9, 2018
Finding a Vector in Cartesian form Mar 22, 2018
Trigonometry - compound angels Jun 8, 2011