Question

Two vectors `vec"A"` and `vec"B"` have equal magnitudes. If magnitude of `vec"A"` + `vec"B"` is equal to two times the magnitude of `vec"A"` - `vec"B"`, then the angle between  `vec"A"` and `vec"B"` will be :

Options

• `sin^-1 (3/5)`

• `sin^-1 (1/3)`

• `cos^-1 (3/5)`

• `cos^-1 (1/3)`

Answer 1

`bb(cos^-1 (3/5))`

Explanation:

Given `|vec"A" +vec"B"| = 2 |vec"A" - vec"B"|`      ...(1)

Let angle between `vec"A"` and `vec"B"` be `theta`.

|A| = |B|

Magnitude of resultant of two vectors is given by

`|vec"R"| = sqrt("A"^2 + "B"^2 + 2"AB" costheta)` 

`|vec"A" + vec"B"| = sqrt("A"^2 + "B"^2 + 2"AB" costheta)` 

`|vec"A" - vec"B"| = sqrt("A"^2 + "B"^2 + 2"AB" cos(pi -theta)` 

= `sqrt("A"^2 + "B"^2 + 2"AB" costheta)` 

Substituting values of `|vec"A" + vec"B"|` and `|vec"A" - vec"B"|` equation (1), we get :

`sqrt("A"^2 + "B"^2 + 2"AB" costheta)` 

= `2sqrt("A"^2 + "B"^2 - 2"AB" costheta)` 

On squaring both the sides

`("A"^2 + "B"^2 + 2"AB" costheta) = 4("A"^2 + "B"^2 - 2"AB" costheta)`

As  |A| = |8| ⇒ A = B

2 + 2 cos θ = 4 (2 – 2 cos θ)

6 = 10 cos θ

 cos θ = `6/10`

∴  cos θ = `3/5` ⇒ θ cos^-1 `(3/5)`