Question

Vector A❝ has components Ax = 1.31cm , Ay = 2.18cm ; vector B❝ has components Bx = 3.99cm , By = -3.78cm .

1 Find the componentes of the Vector A+B.

2 find the magnitude of the vector sum A+ B.

3 find the counterlockwise a angle sum A+ B makes with tbe +x axis.

and how about the vector diference B-A?

Answer 1

1).

A + B = (1.31 + 3.99) i + (2.18 -3.78) j = 5.3 i - 1.6 j

2).

magnitude = sqrt[5.3^2 + 1.6^2] = 5.536 cm

3).
(A+B) lies in the 4th quadrant.

angle with x-axis in anticlockwise sense = 360 - tan^-1(1.6/5.3) = 343.2 degrees

B - A = (3.99 - 1.31) i + (-3.78 - 2.18) j = 2.68 i - 5.96 j

magnitude of (B-A) = sqrt[2.68^2 + 5.96^2] = 6.535 cm

(B-A) lies in the 4th quadrant.

angle with x-axis in anticlockwise sense = 360 - tan^-1(5.96/2.68) = 294.21 degrees

Answer 2

a) x component

1.31 + 3.99 = 5.3

y component
2.18 - 3.78=-1.6

2) mag = sqrt(5.3^2 + 1.6^2)=5.54 cm

3) angle = arctan( -1.6/5.3)=-16.8 degrees = 343.2 degrees

Answer 3

1).

A + B = (1.31 + 3.99) i + (2.18 -3.78) j = 5.3 i - 1.6 j

2).

magnitude = sqrt[5.3^2 + 1.6^2] = 5.536 cm

3).
(A+B) lies in the 4th quadrant.

angle with x-axis in anticlockwise sense = 360 - tan^-1(1.6/5.3) = 343.2 degrees

B - A = (3.99 - 1.31) i + (-3.78 - 2.18) j = 2.68 i - 5.96 j