Let `vecr` be position vector of variable point in cartesian plane OXY such that `vecr*(vecr+6hatj)=7` cuts the co-ordinate axes at four distinct points, then the area of the quadrilateral formed by joining these points is :

Options

A. `4sqrt(7)`
B. `6sqrt(7)`
C. `7sqrt(7)`
D. `8sqrt(7)`

Answer

To solve the problem, we need to find the area of the quadrilateral formed by the points where the equation \( \vec{r} \cdot (\vec{r} + 6\hat{j}) = 7 \) intersects the coordinate axes. Let's break down the solution step by step.

### Step 1: Define the position vector
Let the position vector \( \vec{r} \) be defined as:
\[ \vec{r} = x \hat{i} + y \hat{j} \]
where \( x \) and \( y \) are the coordinates of the point in the Cartesian plane.

### Step 2: Substitute into the equation
Now, substitute \( \vec{r} \) into the given equation:
\[ \vec{r} \cdot (\vec{r} + 6\hat{j}) = 7 \]
This expands to:
\[ (x \hat{i} + y \hat{j}) \cdot (x \hat{i} + (y + 6) \hat{j}) = 7 \]
Calculating the dot product gives:
\[ x^2 + y(y + 6) = 7 \]
This simplifies to:
\[ x^2 + y^2 + 6y = 7 \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:
\[ x^2 + y^2 + 6y - 7 = 0 \]
To complete the square for the \( y \) terms, we add and subtract 9:
\[ x^2 + (y^2 + 6y + 9) - 9 - 7 = 0 \]
This simplifies to:
\[ x^2 + (y + 3)^2 = 16 \]

### Step 4: Identify the circle
The equation \( x^2 + (y + 3)^2 = 16 \) represents a circle with:
- Center at \( (0, -3) \)
- Radius \( 4 \)

### Step 5: Find the intersection points with axes
To find the points where this circle intersects the axes, we set \( y = 0 \) and \( x = 0 \) separately.

#### Intersection with the x-axis (\( y = 0 \)):
\[ x^2 + (0 + 3)^2 = 16 \implies x^2 + 9 = 16 \implies x^2 = 7 \implies x = \pm \sqrt{7} \]
Thus, the points are:
\[ (\sqrt{7}, 0) \quad \text{and} \quad (-\sqrt{7}, 0) \]

#### Intersection with the y-axis (\( x = 0 \)):
\[ 0^2 + (y + 3)^2 = 16 \implies (y + 3)^2 = 16 \implies y + 3 = \pm 4 \]
This gives:
\[ y + 3 = 4 \implies y = 1 \quad \text{and} \quad y + 3 = -4 \implies y = -7 \]
Thus, the points are:
\[ (0, 1) \quad \text{and} \quad (0, -7) \]

### Step 6: Identify the vertices of the quadrilateral
The vertices of the quadrilateral formed by these points are:
- \( A(\sqrt{7}, 0) \)
- \( B(-\sqrt{7}, 0) \)
- \( C(0, 1) \)
- \( D(0, -7) \)

### Step 7: Calculate the area of the quadrilateral
To find the area of quadrilateral \( ABCD \), we can split it into two triangles \( ABC \) and \( ACD \).

#### Area of triangle \( ABC \):
- Base \( AB = 2\sqrt{7} \)
- Height from \( C(0, 1) \) to line \( AB \) is \( 1 \).

Area of triangle \( ABC \):
\[ \text{Area}_{ABC} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2\sqrt{7} \times 1 = \sqrt{7} \]

#### Area of triangle \( ACD \):
- Base \( AD = 7 \)
- Height from \( C(0, -7) \) to line \( AD \) is \( 7 \).

Area of triangle \( ACD \):
\[ \text{Area}_{ACD} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2\sqrt{7} \times 7 = 7\sqrt{7} \]

### Step 8: Total area of quadrilateral \( ABCD \)
\[ \text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD} = \sqrt{7} + 7\sqrt{7} = 8\sqrt{7} \]

### Final Answer
The area of the quadrilateral formed by joining these points is:
\[ \boxed{8\sqrt{7}} \]