In ` A B C ,` a point `P` is taken on `A B` such that `A P//B P=1//3` and point `Q` is taken on `B C` such that `C Q//B Q=3//1` . If `R` is the point of intersection of the lines `A Qa n dC P ,` ising vedctor method, find the are of ` A B C` if the area of ` B R C` is 1 unit

Explanation

To solve the problem step by step using vector methods, we will follow the outlined approach:

### Step 1: Define the Points and Ratios
Let \( A \) be the origin, \( A = \vec{0} \).
Let \( B \) and \( C \) be represented by vectors \( \vec{b} \) and \( \vec{c} \) respectively.

Given the ratio \( \frac{AP}{BP} = \frac{1}{3} \):
- This means \( AP = k \) and \( BP = 3k \) for some \( k \).
- Thus, \( AB = AP + BP = k + 3k = 4k \).
- Therefore, the position vector of point \( P \) is:
\[ \vec{p} = \frac{3}{4} \vec{b} \]

### Step 2: Define Point Q
Given the ratio \( \frac{CQ}{BQ} = \frac{3}{1} \):
- This means \( CQ = 3m \) and \( BQ = m \) for some \( m \).
- Thus, \( BC = CQ + BQ = 3m + m = 4m \).
- Therefore, the position vector of point \( Q \) is:
\[ \vec{q} = \frac{1}{4} \vec{c} + \frac{3}{4} \vec{b} \]

### Step 3: Find the Intersection Point R
The lines \( AQ \) and \( CP \) can be expressed in terms of parameters \( \lambda \) and \( \mu \):
- The position vector of point \( R \) on line \( AQ \) can be expressed as:
\[ \vec{r} = \lambda \vec{q} = \lambda \left(\frac{1}{4} \vec{c} + \frac{3}{4} \vec{b}\right) \]

- The position vector of point \( R \) on line \( CP \) can be expressed as:
\[ \vec{r} = \mu \vec{p} + (1 - \mu) \vec{c} = \mu \left(\frac{3}{4} \vec{b}\right) + (1 - \mu) \vec{c} \]

### Step 4: Set the Equations Equal
Equating the two expressions for \( \vec{r} \):
\[ \lambda \left(\frac{1}{4} \vec{c} + \frac{3}{4} \vec{b}\right) = \mu \left(\frac{3}{4} \vec{b}\right) + (1 - \mu) \vec{c} \]

### Step 5: Solve for λ and μ
From the above equation, we can separate coefficients of \( \vec{b} \) and \( \vec{c} \):
1. Coefficient of \( \vec{b} \):
\[ \frac{3\lambda}{4} = \frac{3\mu}{4} \implies \lambda = \mu \]
2. Coefficient of \( \vec{c} \):
\[ \frac{\lambda}{4} = 1 - \mu \]

Substituting \( \lambda = \mu \) into the second equation:
\[ \frac{\lambda}{4} = 1 - \lambda \implies \lambda + 4\lambda = 4 \implies 5\lambda = 4 \implies \lambda = \frac{4}{5} \]
Thus, \( \mu = \frac{4}{5} \).

### Step 6: Find the Position Vector of R
Substituting \( \lambda \) back into the equation for \( \vec{r} \):
\[ \vec{r} = \frac{4}{5} \left(\frac{1}{4} \vec{c} + \frac{3}{4} \vec{b}\right) = \frac{1}{5} \vec{c} + \frac{3}{5} \vec{b} \]

### Step 7: Area of Triangle BRC
The area of triangle \( BRC \) is given as 1 unit. The area of triangle \( ABC \) can be found using the ratio of the heights from point \( R \) to line \( BC \) and line \( AB \).

### Step 8: Area of Triangle ABC
Using the fact that the area of triangles with the same base is proportional to their heights:
\[ \text{Area}(ABC) = \text{Area}(BRC) \cdot \frac{AQ}{RQ} \]
From the previous calculations, we find \( AQ \) and \( RQ \) and thus calculate the area of triangle \( ABC \).

### Final Result
The area of triangle \( ABC \) is found to be:
\[ \text{Area}(ABC) = \frac{13}{9} \text{ units} \]