Question

Each of the displacement vectors [vector A] and [vector B] shown in the figure below has a magnitude of 3.00 m. Find the following values graphically. Report all angles counterclockwise from the positive x axis.
(a) [vector A] + [vector B]=?

maginitude=?

θ °=?

(b) [vector A] - [vector B]=?

magnitude =?
θ °=?

(c) [vector B] - [vector A]=?

magnitude=?
θ °=?

(d) [vector A] - 2 [vector B]=?

magnitude=?
θ °=?

Answer 1

A = 3 cos30 i + 3 sin30 j
B = 0i + 3 j

(a) [vector A] + [vector B]=?
maginitude=?
θ °=?

[vector A] + [vector B]= (3 cos30 +0)i + (3sin30 +3)j
Magnitude = 4.74

Angle = Arcstan((3sin30 +3)/3cos30)
ANgle = 52.41 Counter clock wise from the x-axis

(b) [vector A] - [vector B]=?
magnitude =?
θ °=?

[vector A] - [vector B]= 3cos30i + (3sin30-3)j
Magnitude = (3cos30^2 + (3sin30-3)^2)^1/2 = 2.12

Angle = arcstan((3sin30-3)/3cos30)
Angle = -23.41 degrees Counter clock wise from the x-axis

(c) [vector B] - [vector A]=?
magnitude=?
θ °=?

[vector B] - [vector A] = - 3cos30i + (3-3sin30)j
magnitude = (-3cos30^2 + (-3sin30+3)^2)^1/2 = 2.12

Angle = arcstan(-(3sin30-3)/-3cos30)
Angle = -23.41 degrees Counter clock wise from the x-axis

(d) [vector A] - 2 [vector B]=?
magnitude=?
θ °=?

[vector A] - 2 [vector B]= 3cos30i + (3sin30-6)j
mAgnitude = (3cos30^2+ (3sin30-6)^2)^.5
Magnitude = 4.74

Angle = arcstan((3sin30-6)/3cos30)
Angle = -52.41 degrees Counter clock wise from the x-axis