__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= 2

**i**+ 3

**j**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

_{x}

^{2}+ P

_{y}

^{2})

^{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) = 45

^{0 }.......(ii)

Hence, the vector

**i**+

**j**makes an angle of 45

^{0}with the x-axis.

Let Q =

**i**+

**j**

Q

_{x}i - Q

_{y}j =

**i**-

**j**

Q

_{x}= Q

_{y}= 1

**|**Q

**|**= (Q

_{x}

^{2}+ Q

_{y}

^{2})

^{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) = -45

^{0}......(iv)

Hence, the vector

**i**-

**j**makes an angle of -45

^{0}with the x-axis.

It is given that:

A = 2

**i**+ 3

**j**

A

_{x}

**i**+ A

_{y}

**j**= 2

**i**+ 3

**j**

On comparing the coefficients of

**i**and

**j**, we have

A

_{x}= 2 and A

_{y}= 3

**|**A

**|**= (2

^{2}+3

^{2})

^{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) = 56.31

^{0}

Angle between the vectors (2

**i**+ 3

**j**) and (

**i**+

**j**), θ' = 56.31 - 45 = 11.31

^{0}

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 (2

**i**+ 3

**j**) and (

**i**-

**j**).

θ" = 45 + 56.31 = 101.31

^{0}

Component of vector A, along the direction of Q, making an angle θ"

= (A Cosθ")Q = (A Cosθ") (

**i**-

**j**) / √2

= √13 Cos (901.31

^{0}) (

**i**-

**j**) / √2

= -0.5 (

**i**-

**j**)

= -5 X √2 / 10

= -1 / √2 ....(vi)

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