A=B =a and angle =theta=¤. A- B = √A^2+B^2- 2ABCos(theta ) = √a^2+a^2 - 2a^2Cos¤ = √2a^2 - 2a^2Cos¤ =√ 2a^2( 1-Cos¤) ......... [1-Cos¤ = 2Sin^2 (¤/2)] =√ 2a^2 *2Sin^2(¤/2) =√4a^2Sin^2(¤/2) = 2aSin(¤/2) IS THE MAGNITUDE.

Introduction:

When two vectors have equal magnitudes and an angle between them, we can find the magnitude of their difference using trigonometry and vector addition/subtraction. In this explanation, we will derive the formula to calculate the magnitude of the difference between two vectors with equal magnitude 'a' and an angle between them denoted by theta.

Derivation:

Let's consider two vectors A and B with equal magnitudes 'a' and an angle between them denoted by theta.

Step 1: Decompose the vectors A and B into their horizontal and vertical components:

- Vector A: Ax = a * cos(theta), Ay = a * sin(theta)
- Vector B: Bx = a * cos(180° - theta) = -a * cos(theta), By = a * sin(180° - theta) = -a * sin(theta)

Step 2: Find the difference between the horizontal and vertical components:

- Difference in horizontal components: Dx = Ax - Bx = a * cos(theta) - (-a * cos(theta)) = 2a * cos(theta)
- Difference in vertical components: Dy = Ay - By = a * sin(theta) - (-a * sin(theta)) = 2a * sin(theta)

Step 3: Use the Pythagorean theorem to find the magnitude of the difference vector:

- Magnitude of the difference vector, D = sqrt(Dx^2 + Dy^2)
- D = sqrt((2a * cos(theta))^2 + (2a * sin(theta))^2)
- D = sqrt(4a^2 * (cos^2(theta) + sin^2(theta)))
- D = sqrt(4a^2) = 2a

Conclusion:

The magnitude of the difference between two vectors with equal magnitudes 'a' and an angle between them denoted by theta is always equal to 2a. This means that the magnitude of the difference vector is independent of the angle between the vectors and only depends on the magnitude of the vectors.