Let a and b be positive real numbers . Suppose `vec(PQ) = a hat(i) + b hat(j) and vec(PS) = a hat(i) - b hat(j) ` are adjacent sides of a parallelofram PQRS . Let `vec(u) and vec(v) ` be the vectors of `vec(w) = hat(i) + hat(j) ` along `vec(PQ) and vec(PS)` respectively . If `|vec(u)| + |vec(v)| = |vec(w)|` and if the area of the parallelogram PQRS is 8 , then which of the following statements is/are TRUE ?

Options

A. `a+b=4`
B. `a-b=2`
C. The length of the diagonal P R of the parallelogram PQRS is 4
D. `vec(w)` is an angle of the vectors ` vec(PQ) and vec(PS)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the vectors
Given:
- \(\vec{PQ} = a \hat{i} + b \hat{j}\)
- \(\vec{PS} = a \hat{i} - b \hat{j}\)

### Step 2: Find the area of the parallelogram
The area \(A\) of the parallelogram formed by vectors \(\vec{PQ}\) and \(\vec{PS}\) can be calculated using the cross product:
\[ A = |\vec{PQ} \times \vec{PS}| \]
Calculating the cross product:
\[ \vec{PQ} \times \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & b & 0 \\ a & -b & 0 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{k} \left( a(-b) - b(a) \right) = \hat{k}(-ab - ab) = -2ab \hat{k} \]
Thus, the area is:
\[ A = | -2ab | = 2ab \]
Given that the area is 8:
\[ 2ab = 8 \implies ab = 4 \]

### Step 3: Find the projections
Let \(\vec{w} = \hat{i} + \hat{j}\). The unit vector along \(\vec{PQ}\) is:
\[ \hat{u} = \frac{\vec{PQ}}{|\vec{PQ}|} = \frac{a \hat{i} + b \hat{j}}{\sqrt{a^2 + b^2}} \]
The projection of \(\vec{w}\) onto \(\vec{PQ}\) is:
\[ \vec{u} = \left( \vec{w} \cdot \hat{u} \right) \hat{u} = \left( \frac{1}{\sqrt{2}} \cdot \frac{a + b}{\sqrt{a^2 + b^2}} \right) \hat{u} \]
Similarly, the unit vector along \(\vec{PS}\) is:
\[ \hat{v} = \frac{\vec{PS}}{|\vec{PS}|} = \frac{a \hat{i} - b \hat{j}}{\sqrt{a^2 + b^2}} \]
The projection of \(\vec{w}\) onto \(\vec{PS}\) is:
\[ \vec{v} = \left( \vec{w} \cdot \hat{v} \right) \hat{v} = \left( \frac{1}{\sqrt{2}} \cdot \frac{a - b}{\sqrt{a^2 + b^2}} \right) \hat{v} \]

### Step 4: Use the given condition
We know that:
\[ |\vec{u}| + |\vec{v}| = |\vec{w}| \]
Calculating the magnitudes:
\[ |\vec{w}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Thus:
\[ |\vec{u}| + |\vec{v}| = \sqrt{2} \]

### Step 5: Solve the equations
From the previous steps, we have:
\[ \frac{a + b}{\sqrt{a^2 + b^2}} + \frac{a - b}{\sqrt{a^2 + b^2}} = \sqrt{2} \]
This simplifies to:
\[ \frac{2a}{\sqrt{a^2 + b^2}} = \sqrt{2} \]
Squaring both sides:
\[ \frac{4a^2}{a^2 + b^2} = 2 \implies 4a^2 = 2(a^2 + b^2) \implies 2a^2 = 2b^2 \implies a^2 = b^2 \]
Since \(a\) and \(b\) are positive, we have \(a = b\).

### Step 6: Find the values of \(a\) and \(b\)
From \(ab = 4\) and \(a = b\):
\[ a^2 = 4 \implies a = 2, b = 2 \]

### Step 7: Check the statements
1. \(a + b = 4\) (True)
2. \(a - b = 0\) (True)
3. Length of diagonal \(PR = 4\) (True)
4. \(w\) is an angle bisector (False)

### Conclusion
The true statements are:
- \(a + b = 4\)
- \(a - b = 0\)
- Length of diagonal \(PR = 4\)