Given that `P=Q=R`. If `vec(P)+vec(Q)=vec(R)` then the angle between `vec(P)` and `vec(R)` is `theta_(1)`. If `vec(P)+vec(Q)+vec(R)=vec(0)` then the angle between `vec(P)` and `vec(R)` is `theta_(2)`. The relation between `theta_(1)` and `theta_(2)` is :-

Options

A. (a)`theta_(1) = theta_(2)`
B. (b)`theta_(1) = (theta_(2))/(2)`
C. (c)`theta_(1) = 2theta_(2)`
D. (d)None of the above

Answer

To solve the problem step by step, we will analyze the two scenarios given in the question and derive the angles θ₁ and θ₂.

### Step 1: Analyze the first condition
We are given that \( \vec{P} + \vec{Q} = \vec{R} \) and \( P = Q = R \).

Since \( P = Q = R \), we can denote them as \( \vec{P} = \vec{Q} = \vec{R} = \vec{x} \).

Thus, the equation becomes:
\[ \vec{x} + \vec{x} = \vec{x} \]
This simplifies to:
\[ 2\vec{x} = \vec{x} \]
This is not possible unless \( \vec{x} = \vec{0} \). However, we will derive the angle \( \theta_1 \) using the cosine rule.

Using the cosine rule:
\[ |\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 - 2|\vec{P}||\vec{Q}|\cos(\theta_1) \]
Substituting \( \vec{P} = \vec{Q} = \vec{R} = \vec{x} \):
\[ |\vec{x}|^2 = |\vec{x}|^2 + |\vec{x}|^2 - 2|\vec{x}||\vec{x}|\cos(\theta_1) \]
This simplifies to:
\[ |\vec{x}|^2 = 2|\vec{x}|^2 - 2|\vec{x}|^2\cos(\theta_1) \]
Rearranging gives:
\[ 2|\vec{x}|^2\cos(\theta_1) = |\vec{x}|^2 \]
Dividing both sides by \( |\vec{x}|^2 \) (assuming \( |\vec{x}| \neq 0 \)):
\[ 2\cos(\theta_1) = 1 \]
Thus:
\[ \cos(\theta_1) = \frac{1}{2} \]
This implies:
\[ \theta_1 = \frac{\pi}{3} \]

### Step 2: Analyze the second condition
Now we consider the second condition where \( \vec{P} + \vec{Q} + \vec{R} = \vec{0} \).

This can be rearranged as:
\[ \vec{P} + \vec{R} = -\vec{Q} \]
Again, substituting \( \vec{P} = \vec{Q} = \vec{R} = \vec{x} \):
\[ \vec{x} + \vec{x} = -\vec{x} \]
This leads to:
\[ 2\vec{x} = -\vec{x} \]
Thus:
\[ 3\vec{x} = 0 \Rightarrow \vec{x} = \vec{0} \]
Now, applying the cosine rule again:
\[ |\vec{Q}|^2 = |\vec{P}|^2 + |\vec{R}|^2 + 2|\vec{P}||\vec{R}|\cos(\theta_2) \]
Substituting \( \vec{P} = \vec{Q} = \vec{R} = \vec{x} \):
\[ |\vec{x}|^2 = |\vec{x}|^2 + |\vec{x}|^2 + 2|\vec{x}||\vec{x}|\cos(\theta_2) \]
This simplifies to:
\[ |\vec{x}|^2 = 2|\vec{x}|^2 + 2|\vec{x}|^2\cos(\theta_2) \]
Rearranging gives:
\[ |\vec{x}|^2 - 2|\vec{x}|^2 = 2|\vec{x}|^2\cos(\theta_2) \]
Thus:
\[ -|\vec{x}|^2 = 2|\vec{x}|^2\cos(\theta_2) \]
Dividing by \( |\vec{x}|^2 \) (assuming \( |\vec{x}| \neq 0 \)):
\[ -1 = 2\cos(\theta_2) \]
This implies:
\[ \cos(\theta_2) = -\frac{1}{2} \]
Thus:
\[ \theta_2 = \frac{2\pi}{3} \]

### Step 3: Relate \( \theta_1 \) and \( \theta_2 \)
Now we have:
\[ \theta_1 = \frac{\pi}{3}, \quad \theta_2 = \frac{2\pi}{3} \]

To find the relation between \( \theta_1 \) and \( \theta_2 \):
\[ \theta_2 = 2\theta_1 \]

### Final Answer
The relation between \( \theta_1 \) and \( \theta_2 \) is:
\[ \theta_2 = 2\theta_1 \]