If `vec(A)=vec(B)+vec(C )`, and the magnitudes of `vec(A)`,`vec(B)`,`vec(C )` are 5,4, and 3 units, then the angle between `vec(A)` and `vec(C )` is

Options

A. `cos^(-1)(3/5)`
B. `cos^(-1)(4/5)`
C. `sin^(-1)(3/4)`
D. `pi/2`

Explanation

To solve the problem, we need to find the angle between vectors A and C given that:

\[ \vec{A} = \vec{B} + \vec{C} \]

and the magnitudes of the vectors are:

- \( |\vec{A}| = 5 \) units
- \( |\vec{B}| = 4 \) units
- \( |\vec{C}| = 3 \) units

### Step 1: Use the Law of Cosines

We can use the Law of Cosines to relate the magnitudes of the vectors and the angle between them. According to the Law of Cosines:

\[ |\vec{A}|^2 = |\vec{B}|^2 + |\vec{C}|^2 + 2 |\vec{B}| |\vec{C}| \cos(\theta) \]

where \( \theta \) is the angle between vectors \( \vec{B} \) and \( \vec{C} \).

### Step 2: Substitute the known values

Substituting the magnitudes into the equation:

\[ 5^2 = 4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos(\theta) \]

Calculating the squares:

\[ 25 = 16 + 9 + 24 \cos(\theta) \]

### Step 3: Simplify the equation

Combine the constants on the right side:

\[ 25 = 25 + 24 \cos(\theta) \]

### Step 4: Isolate the cosine term

Subtract 25 from both sides:

\[ 0 = 24 \cos(\theta) \]

### Step 5: Solve for \( \cos(\theta) \)

From the equation above, we can see that:

\[ \cos(\theta) = 0 \]

### Step 6: Find the angle \( \theta \)

The cosine of an angle is zero at:

\[ \theta = 90^\circ \]

Thus, the angle between vectors \( \vec{A} \) and \( \vec{C} \) is:

\[ \theta = 90^\circ \]

### Final Answer

The angle between \( \vec{A} \) and \( \vec{C} \) is \( 90^\circ \).

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