If the points with position vectors ` 20 hati + p hatj , 5 hati - hatj and 10 hati - 13 hatj` are collinear, then p =

Options

A. 7
B. -37
C. -7
D. 37

Answer

To find the value of \( p \) such that the points with position vectors \( \mathbf{A} = 20 \hat{i} + p \hat{j} \), \( \mathbf{B} = 5 \hat{i} - \hat{j} \), and \( \mathbf{C} = 10 \hat{i} - 13 \hat{j} \) are collinear, we can use the concept that the area of the triangle formed by these three points must be zero.

### Step 1: Set up the area of the triangle
The area \( A \) of the triangle formed by the points \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \) can be calculated using the determinant of the matrix formed by their position vectors:

\[ A = \frac{1}{2} \begin{vmatrix} 1 & 20 & p \\ 1 & 5 & -1 \\ 1 & 10 & -13 \end{vmatrix} \]

### Step 2: Calculate the determinant
We need to calculate the determinant:

\[ \begin{vmatrix} 1 & 20 & p \\ 1 & 5 & -1 \\ 1 & 10 & -13 \end{vmatrix} \]

Expanding this determinant along the first row:

\[ = 1 \cdot \begin{vmatrix} 5 & -1 \\ 10 & -13 \end{vmatrix} - 20 \cdot \begin{vmatrix} 1 & -1 \\ 1 & -13 \end{vmatrix} + p \cdot \begin{vmatrix} 1 & 5 \\ 1 & 10 \end{vmatrix} \]

### Step 3: Calculate the smaller determinants
Now, we calculate the smaller 2x2 determinants:

1. \( \begin{vmatrix} 5 & -1 \\ 10 & -13 \end{vmatrix} = (5)(-13) - (-1)(10) = -65 + 10 = -55 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & -13 \end{vmatrix} = (1)(-13) - (-1)(1) = -13 + 1 = -12 \)

3. \( \begin{vmatrix} 1 & 5 \\ 1 & 10 \end{vmatrix} = (1)(10) - (5)(1) = 10 - 5 = 5 \)

### Step 4: Substitute back into the determinant
Now substituting these values back into the determinant:

\[ = 1(-55) - 20(-12) + p(5) = -55 + 240 + 5p \]

### Step 5: Set the area to zero
Since the points are collinear, the area must be zero:

\[ -55 + 240 + 5p = 0 \]

### Step 6: Solve for \( p \)
Now, we solve for \( p \):

\[ 5p = -240 + 55 \]
\[ 5p = -185 \]
\[ p = \frac{-185}{5} = -37 \]

Thus, the value of \( p \) is \( -37 \).

### Final Answer
\[ p = -37 \]