If Vector A has x and y components of -6.4 and 7.8. Vector B has x and y components of 8.9 and 13. What is th?

If Vector A has x and y components of -6.4 and 7.8. Vector B has x and y components of 8.9 and 13. What is the magnitide of C=A+B

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  • moe
    Lv 7
    9 years ago
    Favorite Answer

    Draw x and y axis. With the origin point as (letter) O (assume)

    Given Ax = -6.4 let OA’ = 6.4 towards LHS of the point O (towards (-) side of x-axis)

    Ay = 7.8 let OA” = 7.8 towards + y-axis (top of the point O) (towards (+) side of y-axis)

    So, resultant vector OA = OA” + A”A :a”A = OA’ = |-6.4| = 6.4

    OA = 7.8+6.4 or

    OA = 14.2

    Magnitude of vector A = 12.2

    Angle A”OA:

    Tan(A”OA) = tanA”A = (p/b) = A”A / OA” = 6.4/7.8 = 0.82

    Tan(theta) = 0.82

    Theta = 61.4 degrees = <A”OA so.

    A”OA’ = 61.4 degrees

    Likewise,

    Bx =OB’=8.9 and By =OB“= 13

    Resultant vector B :

    OB = OBx+OBy = OB’+OB” = 8.9+13=21.9

    Magnitude of vector B = 21.9

    Angle BOB”

    Tan( BOB”) = (p/b) = 8.9/13=0.68 then

    Angle BOB”= 46.33 degrees

    C = A+B = 12.2+21.9 = 34.1 <= answer

  • 3 years ago

    The x and y axes are at top angles to a minimum of one yet another, so the x and y aspects of your vector are the aspects of a top triangle with hypotenuse of four. by way of fact the x part is two times the y part, 4^2 = (2y)^2 + y^2 = 4y^2 + y^2 = 5y^2 sixteen/5 = y^2, so y = 4/sqrt(5) and x is two times that, or 8/sqrt(5)

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