The magnitudes of two vectors `vecP` and `vecQ` differ by 1. The magnitude of their resultant makes an angle of `tan^(-1)(3//4)` with P. The angle between P and Q is

Options

A. `45^(@)`
B. `0^(@)`
C. `180^(@)`
D. `90^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Magnitudes of the Vectors
Let the magnitudes of the vectors \( \vec{P} \) and \( \vec{Q} \) be \( p \) and \( q \), respectively. According to the problem, we know that:
\[ |p - q| = 1 \]
This means that either \( p = q + 1 \) or \( q = p + 1 \).

### Step 2: Use the Angle Information
We are given that the angle \( \theta \) between the resultant vector and vector \( \vec{P} \) is:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]
From the definition of tangent, we can express this as:
\[ \tan \theta = \frac{|\vec{Q}|}{|\vec{P}|} = \frac{q}{p} \]
Thus, we have:
\[ \frac{q}{p} = \frac{3}{4} \]
This implies:
\[ 4q = 3p \quad \text{or} \quad q = \frac{3}{4}p \]

### Step 3: Substitute and Solve for Magnitudes
Now, we can substitute \( q = \frac{3}{4}p \) into the equation \( |p - q| = 1 \):
\[ p - \frac{3}{4}p = 1 \]
This simplifies to:
\[ \frac{1}{4}p = 1 \implies p = 4 \]
Now substituting back to find \( q \):
\[ q = \frac{3}{4} \times 4 = 3 \]

### Step 4: Calculate the Magnitude of the Resultant
Now we can find the magnitude of the resultant \( R \) using the formula:
\[ R = \sqrt{p^2 + q^2 + 2pq \cos \alpha} \]
Since we need to find \( \alpha \), we first calculate \( R \):
\[ R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 5: Use the Resultant Magnitude to Find the Angle
Using the vector addition law:
\[ R^2 = p^2 + q^2 + 2pq \cos \alpha \]
Substituting the known values:
\[ 5^2 = 4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cos \alpha \]
This becomes:
\[ 25 = 16 + 9 + 24 \cos \alpha \]
Simplifying gives:
\[ 25 = 25 + 24 \cos \alpha \]
Thus:
\[ 24 \cos \alpha = 0 \implies \cos \alpha = 0 \]
This means:
\[ \alpha = 90^\circ \]

### Conclusion
The angle between vectors \( \vec{P} \) and \( \vec{Q} \) is \( 90^\circ \).