Question

12.0 m )j . What are (a) the Here are two vectors: а-(6.00 m )i-4.50 m lj and b-| 9.00 m li + magnitude and (b) the angle (counterclockwise from the axis defined by ) of a ? what are (c) the magnitude and (d) the angle of b ? What are (e) the magnitude and (f) the angle of a + b ; (g) the magnitude and (h) the angle of b-a and (i) the magnitude and () the angle of a -b (State your angle as a positive number.) (k) What is the angle between the directions of b -a and a-b Units (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (9) Number (h) Number (i) Number Units Units Units Units Units Units Units Units 6i) Number Units

Answer 1

given

a = 6.00 i - 4.50 j

b = 9.00 i + 12.0 j

a )

| a | = ( 6² + (-4.5)² )^1/2

= 7.5 m

b )

angle = tan^-1 ( -4.5/6 )

= - 36.87^o

here 360 - 36.87 = 323.13^o anti clock wise from the x - axis

c )

| b | = ( 9² + 12² )^1/2

= 15 m

d )

angle = tan^-1 ( 12/9 )

= 53.13^o

here 53.13^o anti clock wise from the x - axis

e )

a + b = ( 6 + 9 ) i + ( -4.5 + 12 ) j

= 15 i + 7.5 j

| a + b | = ( 15² + 7.5² )^1/2

= 16.77 m

f )

angle is = tan^-1 ( 7.5/15 )

= 26.56^o

here 26.56^o anti clock wise from the x - axis

g )

| b - a | = | ( 6 - 9 ) i | + | ( 12 + 4.5 )j |

= 3 i + 16.5 j

| b - a | = ( 3² + 16.5² )^1/2

= 16.77 m

h )

angle = tan^-1 ( 16.5/3 )

= 79.695^o anti clock wise from x - axis

i )

a - b = - 3 i - 16.5 j

| a - b | = 16.77 m

j )

angle = tan^-1 ( - 16.5 / -3 )

= 79.695^o

it is now 360 - 79.695 = 259.695^o

k )

so the angle is = 259.695 - 79.695

= 180^o