Question

iˆ and jˆ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors iˆ + jˆ, and iˆ − j ˆ? What are the components of a vector A= 2 iˆ + 3j ˆalong the directions of iˆ + j ˆand iˆ − jˆ? [You may use graphical method]

Answer 1

Consider a vector p, given as,

p= i ^ + j ^

The magnitude of a vector, p= p x i ^ + p y j ^ is given as,

p= ( p x ) 2 + ( p y ) 2

By substituting the given values in the above expression, we get

| p |= ( 1 ) 2 + ( 1 ) 2 = 2

The direction of the vector p is given as,

θ= tan −1 p y p x

By substituting the given values in the above expression, we get

θ= tan −1 ( 1 1 ) = tan −1 ( 1 ) =45°

Thus, the magnitude of p is 2 unitsand its direction is 45°above the x –axis.

Consider a vector q, given as,

q= i ^ − j ^

The magnitude of a vector qis,

| q |= ( 1 ) 2 + ( −1 ) 2 = 2

The direction of the vector qis,

β= tan −1 ( −1 1 ) = tan −1 ( −1 ) =−45°

Thus, the magnitude of qis 2 unitsand its direction is −45°with the x–axis.

The vector A is given as A=2 i ^ +3 j ^ .

The magnitude of Ais,

| A |= ( 3 ) 2 + ( 2 ) 2 = 13

The direction of the vector Aalong x–axis is given as,

ϕ= tan −1 ( 3 2 ) = tan −1 ( 1.5 ) =56.31°

The angle between the vectors A and p is given as,

γ=ϕ−θ

By substituting the given values in the above expression, we get

γ=56.31°−45° =11.31°

The component of Aalong p is given as,

( Acosθ ) p | p | = 13 cos( 11.31° )( i ^ + j ^ 2 ) =2.5( i ^ + j ^ ) =2.5 2 = 5 2

The angle between the vectors A and q is given as,

α=ϕ−β

By substituting the given values in the above expression, we get

α=56.31°−( −45° ) =101.31°

Component of Aalong qis given as,

( Acos θ ′ ) q | q | = 13 cos( 101.31 )( i ^ − j ^ 2 ) =− 5 10 ( i ^ − j ^ ) =− 1 2 2 =− 1 2

Thus, the component of vector A along p is 5 2 and its component along q is − 1 2 .