Question

Given vector A bar = 2i^+3j^, the angle between A bar and Y-axis is?

Answer 1

vector A bar = 2i^+3j^

therefore A = |vector A bar| = √(2²+3²) = √13

let β be the angle between vector A and Y axis,

Then cosβ = y/A = 3/√13 and [ where Y co-ordinate y = 3 ]

sin β = x/A = 2/√13 [ as X co-ordinate x = 2

tanβ = x/y = 2/3

Hence, β = tan^-1 (2/3) Ans.