Motion in a plane NCERT Solutions Class 11 Physics - Solved Exercise Question 4.22

Question 4.22:

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

Solution:

Consider a vector P, given as:
P = i + j
P_x i + P_y j = i + j

On comparing the components on both sides, we get:
P_x = P_y = 1
|P| = ( P_x² + P_y² )^1/2
= (1+1)^1/2
= √2    ....(i) 
Hence, the magnitude of the vector i + j is √2
Let θ be the angle made by the vector P, with the x-axis, as shown in the following figure.

∴ tan θ = ( P_y / P_x )
θ = tan^-1 (1 / 1) = 45⁰ .......(ii)
Hence, the vector i + j makes an angle of 45⁰ with the x-axis.
Let Q = i + j
Q_x i - Q_y j = i - j
Q_x = Q_y = 1
|Q|  = (Q_x² + Q_y²)^1/2 = √2     .... (iii)
Hence, the magnitude of the vector i - j is √2
Let θ be the angle made by the vector |Q|, with the x- axis, as shown in the following figure.

∴ tan θ = (-Q_y / Q_x)
θ = tan^-1 (-1/1) = -45⁰     ......(iv)
Hence, the vector i - j makes an angle of -45⁰ with the x-axis.
It is given that:
A = 2i + 3j
A_x i + A_y j = 2i + 3j
On comparing the coefficients of i and j, we have
A_x = 2 and A_y = 3
|A| = (2²+3²)^1/2 = (13)^1/2
Let A_x make an angle θ with the x-axis, as shown in the following figure.

∴ tan θ = (A_y / A_x)
θ = tan^-1 (-3/2)  =  56.31⁰
Angle between the vectors (2i + 3j) and (i + j), θ' = 56.31 - 45 = 11.31⁰
Component of vector A, along the direction of P, making an angle θ'
= (A Cos θ') P^{^} = (A Cos 11.31) (i + j) / √2
= √13 X 0.9806 (i + j) / √2
= 2.5 (i + j)
= 5 / √2   ........(v)
Let θ" be the angle between the vectors (2i + 3j) and (i - j).
θ" = 45 + 56.31 = 101.31⁰
Component of vector A, along the direction of Q, making an angle θ"
= (A Cosθ")Q = (A Cosθ") (i - j) / √2
= √13 Cos (901.31⁰) (i - j) / √2
= -0.5 (i - j)
= -5 X √2 / 10
= -1 / √2     ....(vi)