If A=B+C have scalar magnitudes o 5,4,3 units respectively , then find the between A and C.

Options

A. `cos^(-1)((2)/(5))`
B. `cos^(-1)((1)/(5))`
C. `cos^(-1)((3)/(5))`
D. `sin^(-1)((3)/(5))`

Explanation

To find the angle between vectors A and C given that the magnitudes of A, B, and C are 5, 4, and 3 units respectively, we can use the cosine rule in the context of vector addition. Here’s a step-by-step solution:

### Step 1: Understand the Vector Relationship
We know that the vectors A, B, and C satisfy the equation:
\[ A = B + C \]
Given the magnitudes:
- \( |A| = 5 \)
- \( |B| = 4 \)
- \( |C| = 3 \)

### Step 2: Apply the Cosine Rule
In a triangle formed by vectors A, B, and C, we can apply the cosine rule:
\[ |A|^2 = |B|^2 + |C|^2 + 2|B||C|\cos(\theta) \]
Where \( \theta \) is the angle between vectors B and C.

### Step 3: Substitute the Known Values
Substituting the magnitudes into the cosine rule:
\[ 5^2 = 4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos(\theta) \]
This simplifies to:
\[ 25 = 16 + 9 + 24\cos(\theta) \]

### Step 4: Simplify the Equation
Combine the constants on the right side:
\[ 25 = 25 + 24\cos(\theta) \]
Subtract 25 from both sides:
\[ 0 = 24\cos(\theta) \]

### Step 5: Solve for Cosine
Dividing both sides by 24 gives:
\[ \cos(\theta) = 0 \]

### Step 6: Find the Angle
The cosine of an angle is zero at:
\[ \theta = 90^\circ \]

### Conclusion
Thus, the angle between vectors A and C is:
\[ \theta = 90^\circ \]

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