Question

1. Let AB be a vector in two dimensional plane with magnitude 4 units, and making an angle of 60'with
x-axis and lying in first quadrant. Find the components of AB along x axis and y axis. Hence represent
AB in terms of unit vectors î and ſ.​

Answer 1

Answer:

AB = 2 i + 2√3 j

Step-by-step explanation:

Data:

Vector = AB

Angle = α = 60°

Magnitude of vector AB = ║AB║= 4 units

Required:

Component of AB along x-axis = ABx

Component of AB along y-axis = ABy

Calculation:

We know that

    Cos(α) = Base / hypotenuse = ║ABx║ / ║AB║

⇒  Cos(α) × ║AB║= ║ABx║

Putting values we get

     ║ABx║ = Cos(60) × 4

     ║ABx║ = ( 1 / 2 ) × 4 = 2 units

     ║ABx║ = 2

⇒        ABx = 2 i

NOW

     Sin(α) = perpendicular / hypotenuse = ║ABy║ / ║AB║

⇒   Sin(α) × ║AB║ = ║ABy║

Putting values we get

      Sin(60) × 4 = ║ABy║  

⇒         ║ABy║ = Sin(60) × 4

⇒         ║ABy║ = ( √3 / 2 ) × 4

⇒         ║ABy║ = ( √3 ) × 2 =2√3

⇒         ║ABy║ = 2√3

⇒              ABy = 2√3 j  

Thus

AB = ABx + ABy = 2 i + 2√3 j

So

AB = 2 i + 2√3 j