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Vector Addition

  1. Apr 3, 2005 #1
    Q. All four of the vectors below are added to vector E. Positive angles are measured counter-clockwise from the x axis.
    A = 2 m î + 3 m ĵ
    B = -7 m î -10 m ĵ
    C = 3 m at 62 °
    D = 8 m at -226 °

    a) What is C·D?

    C.D = |C| |D| cos theta

    I did this.

    3* cos 62 + 8 cos -226 , but the answer is wrong

    b) Vector F is the cross product of vectors A and B (= A X B).

    What is the z component of vector F?

    The answer is one, but how

    F = A X B

    Find the cross product by evaluating the determinant of vectors A & B.

    Then what!!
  2. jcsd
  3. Apr 3, 2005 #2


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    a) What you did is you substracted the x-component of each vectors. That's not the dot product. The dot product is the product of lenght of the two vectors times the cosine of the shortest angle between then.

    b) According the right-hand rule or screw rule (or whatever rule you are comfortable with), the vector F has ONLY a z component. So all you got to do is find its lenght.
    Last edited: Apr 4, 2005
  4. Apr 3, 2005 #3
    a) The formula [tex]\vec{c} \cdot \vec{d} = |\vec{c}| |\vec{d}| \cos \theta[/tex] takes the cosine of the angle between the two vectors and multiplies the result by the magnitudes of the two vectors.

    The angle between the two vectors is (-226° + 360°) - 62° = 134° - 62° = 72° = [tex]\frac{2\pi}{5}[/tex]

    The magnitude is calculated by adding the square of the components and taking the square root of that sum:

    [tex]|\vec{a}| = \sqrt{x^2 + y^2}[/tex]
    Last edited: Apr 3, 2005
  5. Apr 4, 2005 #4


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    But notice Naeem that for the vectors C and D the magnitudes are already given to you: 3m and 8m.
  6. Apr 4, 2005 #5
    Got, it thanks!
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