Question 119: Motion in 2D & Vectors

Question

Two vectors of same magnitude have a resultant equal to either of the two vectors. The angle between two vectors is

Accepted Answer

Let's consider the two vectors to be $ \mathbf{A} $ and $ \mathbf{B} $, each with magnitude $ a $. According to the question, the magnitude of the resultant vector is equal to the magnitude of either of the two vectors, i.e., resultant vector $ R = a $.
To calculate the angle between the two vectors, we can use the cosine rule for vectors, which gives us the magnitude of the resultant vector $ R $ in terms of the magnitudes of $ \mathbf{A} $ and $ \mathbf{B} $ and the angle $ 	heta $ between them:
$R^2 = A^2 + B^2 + 2AB\cos	heta$
Given that $ A = B = a $ and $ R = a $, our equation becomes:
$a^2 = a^2 + a^2 + 2a \cdot a \cdot \cos	heta$
Simplifying it, we get:
$a^2 = 2a^2 + 2a^2\cos	heta$
$0 = a^2 + 2a^2\cos	heta$
$-1 = 2\cos	heta$
Solving for $ \cos	heta $:
$\cos	heta = -\frac{1}{2}$
The angle $ 	heta $ whose cosine is $ -\frac{1}{2} $ is $ \cos^{-1}(-0.5) $. Therefore, the correct answer is:
Option A: $$\cos^{-1}(-0.5)$$

Answer

$$\cos ^{-1}(-0.5)$$

Answer

$$\cos ^{-1}(-0.4)$$

Answer

$$\cos ^{-1}(-0.3)$$

Answer

$$\cos ^{-1}(-0.6)$$

Motion in 2D & Vectors

Two vectors of same magnitude have a resultant equal to either of the two vectors. The angle between two vectors is

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