If `overset(rarr)A=overset(rarr)B+overset(rarr)C` and the magnitude of `overset(rarr)A, overset(rarr)B` and `overset(rarr)C` are 5,4 and 3 units respectively the angle between `overset(rarr)A` and `overset(rarr)C` is

Options

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

Explanation

To solve the problem, we start with the vector equation given:

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

We know the magnitudes of the vectors:
- \(|\vec{A}| = 5\) units
- \(|\vec{B}| = 4\) units
- \(|\vec{C}| = 3\) units

We need to find the angle between \(\vec{A}\) and \(\vec{C}\).

### Step 1: Use the Law of Cosines

From the vector equation, we can square both sides:

\[ |\vec{A}|^2 = |\vec{B} + \vec{C}|^2 \]

Using the Law of Cosines, we can expand the right-hand side:

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

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

### Step 2: Substitute the Magnitudes

Substituting the known 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 terms on the right:

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

Subtract 25 from both sides:

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

### Step 4: Solve for \(\cos(\theta)\)

From the equation above, we find:

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

### Step 5: Determine the Angle

The angle \(\theta\) for which \(\cos(\theta) = 0\) is:

\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{(or 90 degrees)} \]

This means that vectors \(\vec{B}\) and \(\vec{C}\) are perpendicular to each other.

### Step 6: Find the Angle Between \(\vec{A}\) and \(\vec{C}\)

To find the angle between \(\vec{A}\) and \(\vec{C}\), we can use the cosine rule again.

Let \(\phi\) be the angle between \(\vec{A}\) and \(\vec{C}\):

Using the triangle formed by \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\):

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

Substituting the values:

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

Calculating the squares:

\[ 9 = 25 + 16 - 40 \cos(\phi) \]

Combine the terms:

\[ 9 = 41 - 40 \cos(\phi) \]

Rearranging gives:

\[ 40 \cos(\phi) = 41 - 9 \]

\[ 40 \cos(\phi) = 32 \]

### Step 7: Solve for \(\cos(\phi)\)

\[ \cos(\phi) = \frac{32}{40} = \frac{4}{5} \]

### Step 8: Find the Angle \(\phi\)

Now, we can find \(\phi\):

\[ \phi = \cos^{-1}\left(\frac{4}{5}\right) \]

This gives us the angle between \(\vec{A}\) and \(\vec{C}\).

### Final Answer

The angle between \(\vec{A}\) and \(\vec{C}\) is \(\cos^{-1}\left(\frac{4}{5}\right)\).

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