Question

If `sqrt("A"^2+"B"^2)` represents the magnitude of resultant of two vectors `(vec"A" + vec"B")` and `(vec"A" - vec"B")`, then the angle between two vectors is ______.

Options

• `cos^-1[(2("A"^2-"B"^2))/("A"^2+"B"^2)]`

• `cos^-1[-("A"^2-"B"^2)/("A"^2"B"^2)]`

• `cos^-1[-("A"^2+"B"^2)/(2("A"^2-"B"^2))]`

• `cos^-1[-("A"^2-"B"^2)/("A"^2+"B"^2)]`

Answer 1

If `sqrt("A"^2+"B"^2)` represents the magnitude of resultant of two vectors `(vec"A" + vec"B")` and `(vec"A" - vec"B")`, then the angle between two vectors is `underlinebb(cos^-1[-(A^2+B^2)/(2(A^2-B^2))])`.

Explanation:

As we know that the magnitude of the resultant of two vectors X and Y,

R² = X² + Y² + 2XY cosθ   ...(i)

where, θ is the angle between X and Y.

Putting, X = (A + B)

Y = (A - B)

and R = `sqrt("A"^2 + "B"^2)`  In Eq. (i), we get

A² + B² = (A + 8)² + (A - 8)² + 2(A + 8)(A - 8)cosθ

⇒ A² + B² = A² + B² + 2AB + A² + B² - 2AB + 2(A² - B²)cosθ

`=> (-("A"^2+"B"^2))/(2("A"^2-"B"^2))=costheta`

we get,

`theta=cos^-1[-("A"^2+"B"^2)/(2("A"^2-"B"^2))]`