Three vectors `vec(P) , vec(Q)` and `vec( R)` are such that `|vec(P)| , |vec(Q )|, |vec(R )| = sqrt(2) |vec(P)|` and `vec(P) + vec(Q) + vec(R ) = 0`. The angle between `vec(P)` and `vec(Q) , vec(Q)` and `vec(R )` and `vec(P)` and `vec(R )`are

Options

A. `90^(@) , 135^(@) , 135^(@)`
B. `90^(@) , 45^(@) , 45^(@)`
C. `45^(@) , 90^(@) , 90^(@)`
D. `45^(@) , 135^(@) , 135^(@)`

Explanation

To solve the problem, we need to find the angles between three vectors \( \vec{P}, \vec{Q}, \) and \( \vec{R} \) given the conditions:

1. \( |\vec{P}| = |\vec{Q}| \)
2. \( |\vec{R}| = \sqrt{2} |\vec{P}| \)
3. \( \vec{P} + \vec{Q} + \vec{R} = 0 \)

### Step 1: Assign Magnitudes
Let the magnitude of \( \vec{P} \) be \( x \). Then, the magnitudes of the vectors can be defined as:
- \( |\vec{P}| = x \)
- \( |\vec{Q}| = x \)
- \( |\vec{R}| = \sqrt{2} x \)

### Step 2: Rearranging the Vector Equation
From the equation \( \vec{P} + \vec{Q} + \vec{R} = 0 \), we can rearrange it to:
\[ \vec{P} + \vec{Q} = -\vec{R} \]

### Step 3: Squaring Both Sides
Now, we square both sides of the equation:
\[ |\vec{P} + \vec{Q}|^2 = |\vec{R}|^2 \]
Using the property of magnitudes, this expands to:
\[ |\vec{P}|^2 + |\vec{Q}|^2 + 2 \vec{P} \cdot \vec{Q} = |\vec{R}|^2 \]

### Step 4: Substitute Magnitudes
Substituting the magnitudes we defined:
\[ x^2 + x^2 + 2 \vec{P} \cdot \vec{Q} = (\sqrt{2} x)^2 \]
This simplifies to:
\[ 2x^2 + 2 \vec{P} \cdot \vec{Q} = 2x^2 \]

### Step 5: Solve for the Dot Product
From the equation, we can isolate the dot product:
\[ 2 \vec{P} \cdot \vec{Q} = 0 \]
This implies:
\[ \vec{P} \cdot \vec{Q} = 0 \]
Thus, the angle \( \theta_{PQ} \) between \( \vec{P} \) and \( \vec{Q} \) is:
\[ \theta_{PQ} = 90^\circ \]

### Step 6: Finding the Angle Between \( \vec{Q} \) and \( \vec{R} \)
We apply the same method for \( \vec{Q} + \vec{R} = -\vec{P} \):
\[ |\vec{Q} + \vec{R}|^2 = |\vec{P}|^2 \]
Expanding gives:
\[ |\vec{Q}|^2 + |\vec{R}|^2 + 2 \vec{Q} \cdot \vec{R} = |\vec{P}|^2 \]
Substituting the magnitudes:
\[ x^2 + 2x^2 + 2 \vec{Q} \cdot \vec{R} = x^2 \]
This simplifies to:
\[ 3x^2 + 2 \vec{Q} \cdot \vec{R} = x^2 \]
Isolating the dot product:
\[ 2 \vec{Q} \cdot \vec{R} = -2x^2 \]
Thus:
\[ \vec{Q} \cdot \vec{R} = -x^2 \]
The angle \( \theta_{QR} \) can be found using:
\[ |\vec{Q}||\vec{R}|\cos(\theta_{QR}) = -x^2 \]
Substituting the magnitudes:
\[ x(\sqrt{2} x) \cos(\theta_{QR}) = -x^2 \]
This simplifies to:
\[ \sqrt{2} x^2 \cos(\theta_{QR}) = -x^2 \]
Dividing both sides by \( x^2 \) (assuming \( x \neq 0 \)):
\[ \sqrt{2} \cos(\theta_{QR}) = -1 \]
Thus:
\[ \cos(\theta_{QR}) = -\frac{1}{\sqrt{2}} \]
Therefore:
\[ \theta_{QR} = 135^\circ \]

### Step 7: Finding the Angle Between \( \vec{P} \) and \( \vec{R} \)
Using \( \vec{P} + \vec{R} = -\vec{Q} \):
\[ |\vec{P} + \vec{R}|^2 = |\vec{Q}|^2 \]
Expanding gives:
\[ |\vec{P}|^2 + |\vec{R}|^2 + 2 \vec{P} \cdot \vec{R} = |\vec{Q}|^2 \]
Substituting the magnitudes:
\[ x^2 + 2x^2 + 2 \vec{P} \cdot \vec{R} = x^2 \]
This simplifies to:
\[ 3x^2 + 2 \vec{P} \cdot \vec{R} = x^2 \]
Isolating the dot product:
\[ 2 \vec{P} \cdot \vec{R} = -2x^2 \]
Thus:
\[ \vec{P} \cdot \vec{R} = -x^2 \]
Following the same steps as before, we find:
\[ \theta_{PR} = 135^\circ \]

### Final Angles
The angles between the vectors are:
- \( \theta_{PQ} = 90^\circ \)
- \( \theta_{QR} = 135^\circ \)
- \( \theta_{PR} = 135^\circ \)

### Summary of Results
Thus, the angles between the vectors are:
- Angle between \( \vec{P} \) and \( \vec{Q} \): \( 90^\circ \)
- Angle between \( \vec{Q} \) and \( \vec{R} \): \( 135^\circ \)
- Angle between \( \vec{P} \) and \( \vec{R} \): \( 135^\circ \)