If the vectors `3hat(i)-5hat(j)+hat(k)and9hat(i)-15hat(j)+phat(k)` are collinear, then find the value p.

Answer

To determine the value of \( p \) such that the vectors \( \mathbf{a} = 3\hat{i} - 5\hat{j} + \hat{k} \) and \( \mathbf{b} = 9\hat{i} - 15\hat{j} + p\hat{k} \) are collinear, we can use the property that two vectors are collinear if their cross product is zero.

### Step 1: Set up the cross product
The cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors.

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 1 \\ 9 & -15 & p \end{vmatrix} \]

### Step 2: Calculate the determinant
We can expand this determinant using the first row:

\[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -5 & 1 \\ -15 & p \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 9 & p \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -5 \\ 9 & -15 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -5 & 1 \\ -15 & p \end{vmatrix} = (-5)p - (1)(-15) = -5p + 15 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ 9 & p \end{vmatrix} = (3)p - (1)(9) = 3p - 9 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -5 \\ 9 & -15 \end{vmatrix} = (3)(-15) - (-5)(9) = -45 + 45 = 0 \]

### Step 3: Combine results
Putting it all together, we have:

\[ \mathbf{a} \times \mathbf{b} = (-5p + 15)\hat{i} - (3p - 9)\hat{j} + 0\hat{k} \]

### Step 4: Set the cross product to zero
For the vectors to be collinear, the cross product must equal zero:

\[ (-5p + 15)\hat{i} - (3p - 9)\hat{j} + 0\hat{k} = 0 \]

This gives us two equations:

1. \( -5p + 15 = 0 \)
2. \( 3p - 9 = 0 \)

### Step 5: Solve for \( p \)
From the first equation:

\[ -5p + 15 = 0 \implies 5p = 15 \implies p = 3 \]

From the second equation:

\[ 3p - 9 = 0 \implies 3p = 9 \implies p = 3 \]

Both equations yield the same result.

### Conclusion
Thus, the value of \( p \) is:

\[ \boxed{3} \]