Given that P 12, Q = 5 and R = 13 also P+Q=R, then the angle between P and Q will be

Options

A. `pi`
B. `(pi)/(2)`
C. zero
D. `(pi)/(4)`

Explanation

To find the angle between vectors P and Q given that P = 12, Q = 5, and R = 13, and that P + Q = R, we can use the cosine rule for vectors. The cosine rule states that for any two vectors A and B, the magnitude of their resultant R can be expressed as:

\[ R^2 = A^2 + B^2 + 2AB \cos(\theta) \]

where \( \theta \) is the angle between the vectors A and B.

### Step-by-step Solution:

1. **Identify the given values**:
- \( P = 12 \)
- \( Q = 5 \)
- \( R = 13 \)

2. **Write the cosine rule equation**:
\[ R^2 = P^2 + Q^2 + 2PQ \cos(\theta) \]

3. **Substitute the known values into the equation**:
\[ 13^2 = 12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cos(\theta) \]

4. **Calculate the squares**:
\[ 169 = 144 + 25 + 120 \cos(\theta) \]

5. **Combine the constant terms**:
\[ 169 = 169 + 120 \cos(\theta) \]

6. **Isolate the cosine term**:
\[ 169 - 169 = 120 \cos(\theta) \]
\[ 0 = 120 \cos(\theta) \]

7. **Solve for \( \cos(\theta) \)**:
\[ \cos(\theta) = 0 \]

8. **Determine the angle \( \theta \)**:
The angle whose cosine is 0 is:
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \]

### Final Answer:
The angle between vectors P and Q is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

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