# Plz help..Two vectors A and B have values 27N and 61N, respectively.?

If vector A is at an angle of 51 degrees and vector B is at an angle of 84 degrees.What is the vector difference of the two vectors?

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*PHYSICS question*

### 2 Answers

- VamanLv 74 weeks ago
Magnitude remain the same.=27+61=88. Now you need to find the resulting angle. The angle betweem A and B is 84-51=33. The magnitude of the resultant will be

(vec A-vec B)^2= A^2+B^2-2AB cos 33=r^2

r^2=1687.4 r=41

The angle of the resultant is given by

27= 41 cos a, cos a=27/41 a=48.81degrees

wrt x axis the resultant makes an angle 51+48.81=99.81.

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- 4 weeks ago
Ax = 27 * cos(51)

Ay = 27 * sin(51)

Bx = 61 * cos(84)

By = 61 * sin(84)

A - B

Ax - Bx = 27 * cos(51) - 61 * cos(84)

Ay - By = 27 * sin(51) - 61 * sin(84)

m = sqrt((Ax - Bx)^2 + (Ay - By)^2)

m^2 =>

(Ax - Bx)^2 + (Ay - By)^2 =>

(27 * cos(51) - 61 * cos(84))^2 + (27 * sin(51) - 61 * sin(84))^2 =>

729 * cos(51)^2 - 3294 * cos(51) * cos(84) + 3721 * cos(84)^2 + 729 * sin(51)^2 - 3294 * sin(51) * sin(84) + 3721 * sin(84)^2 =>

729 * (cos(51)^2 + sin(51)^2) + 3721 * (cos(84)^2 + sin(84)^2) - 3294 * (cos(84) * cos(51) + sin(84) * sin(51)) =>

729 * 1 + 3721 * 1 - 3294 * cos(84 - 51) =>

729 + 3721 - 3294 * cos(33) =>

4450 - 3294 * cos(33)

m = sqrt(4450 - 3294 * cos(33))

m * cos(t) = Ax - Bx

m * cos(t) = 27 * cos(51) - 61 * cos(84))

cos(t) = (27 * cos(51) - 61 * cos(84)) / sqrt(4450 - 3294 * cos(33))

t = +/- 75 degrees

m * sin(t) = Ay - By

sin(t) = (27 * sin(51) - 61 * sin(84)) / sqrt(4450 - 3294 * cos(33))

t = -75 , 255

t = -75

m = 41.078207716352149123906185623641

41 Newtons at -75 degrees

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