Solveeit Logo
Login

Question

Question: If |$\vec{A}-\vec{B}$|=|$\vec{A}$|=|$\vec{B}$|, the angle between $\vec{A}$ and $\vec{B}$ is...

If |A⃗−B⃗\vec{A}-\vec{B}|=|A⃗\vec{A}|=|B⃗\vec{B}|, the angle between A⃗\vec{A} and B⃗\vec{B} is

A

0°

B

30°

C

60°

D

90°

Answer

60°

Explanation

Solution

Let ∣A⃗∣=∣B⃗∣=k|\vec{A}| = |\vec{B}| = k. The given condition is ∣A⃗−B⃗∣=k|\vec{A}-\vec{B}| = k. The magnitude of the difference of two vectors is given by:

∣A⃗−B⃗∣2=(A⃗−B⃗)⋅(A⃗−B⃗)=A⃗⋅A⃗−2(A⃗⋅B⃗)+B⃗⋅B⃗|\vec{A}-\vec{B}|^2 = (\vec{A}-\vec{B}) \cdot (\vec{A}-\vec{B}) = \vec{A} \cdot \vec{A} - 2(\vec{A} \cdot \vec{B}) + \vec{B} \cdot \vec{B} ∣A⃗−B⃗∣2=∣A⃗∣2+∣B⃗∣2−2(A⃗⋅B⃗)|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2(\vec{A} \cdot \vec{B})

Let θ\theta be the angle between A⃗\vec{A} and B⃗\vec{B}. The dot product is A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta. Substituting the given conditions into the equation:

k2=k2+k2−2(k⋅kcos⁡θ)k^2 = k^2 + k^2 - 2(k \cdot k \cos\theta) k2=2k2−2k2cos⁡θk^2 = 2k^2 - 2k^2 \cos\theta

Assuming k≠0k \neq 0, we can divide the equation by k2k^2:

1=2−2cos⁡θ1 = 2 - 2 \cos\theta 2cos⁡θ=2−12 \cos\theta = 2 - 1 2cos⁡θ=12 \cos\theta = 1 cos⁡θ=12\cos\theta = \frac{1}{2}

The angle θ\theta between two vectors is conventionally taken in the range [0,π][0, \pi]. In this range, the angle whose cosine is 1/21/2 is θ=π3\theta = \frac{\pi}{3} or 60∘60^\circ.