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# 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

### 2 Answers

- moeLv 79 years agoFavorite 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

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- buchholtzLv 43 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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