High School

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Let \(\vec{B} = 5.15 \, \text{m}\) at \(60.0^\circ\).

Vectors \(\vec{C}\) and \(\vec{A}\) have equal magnitudes. The direction angle of \(\vec{C}\) is larger than that of \(\vec{A}\) by \(25.0^\circ\).

Given:
\[
\vec{A} \cdot \vec{B} = 27.0 \, \text{m}^2
\]
\[
\vec{B} \cdot \vec{C} = 36.3 \, \text{m}^2
\]

What is the magnitude of \(\vec{A}\)?

Answer :

The magnitude of vec(A) is 9.0 m

Given:

- B = 5.15 m at 60.0°

- A and C have equal magnitudes

- The direction angle of C is larger than that of A by 25.0°

- A ⋅ B = 27.0 m²

- B ⋅ C = 36.3 m²

Step 1: Find Components of B

- Bₓ = B ⋅ cos(θ)

= 5.15 m ⋅ cos(60.0°)

= 5.15 m ⋅ 0.5

= 2.575 m

- Bᵧ = B ⋅ sin(θ)

= 5.15 m ⋅ sin(60.0°)

= 5.15 m ⋅ √3 / 2

≈ 4.467 m

Step 2: Find the Angle Between A and B

- A ⋅ B = |A| ⋅ |B| ⋅ cos(θₐᵦ)

- 27.0 m² = |A| ⋅ 5.15 m ⋅ cos(60.0°)

- Solve for |A|:

|A| = 27.0 m² / (5.15 m ⋅ cos(60.0°))

≈ 9.0 m

Step 3: Find the Magnitude of C

- Since A and C have equal magnitudes and the direction angle of C is larger than that of A by 25.0°, the magnitude of C is also 9.0 m.

Therefore, the final answer is |A| = 9.0 m.

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