Find a vector of magnitude 15, which is perpendicular to both the vectors `(4hat(i) -hat(j)+8hat(k)) and (-hat(j)+hat(k)).`

Answer

To find a vector of magnitude 15 that is perpendicular to both vectors \( \mathbf{A} = 4\hat{i} - \hat{j} + 8\hat{k} \) and \( \mathbf{B} = -\hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Define the vectors
Let:
\[ \mathbf{A} = 4\hat{i} - \hat{j} + 8\hat{k} \]
\[ \mathbf{B} = -\hat{j} + \hat{k} \]

### Step 2: Find the cross product \( \mathbf{A} \times \mathbf{B} \)
The cross product of two vectors gives a vector that is perpendicular to both. We can calculate \( \mathbf{A} \times \mathbf{B} \) using the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 8 \\ 0 & -1 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -1 & 8 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 8 \\ 0 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & -1 \\ 0 & -1 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} -1 & 8 \\ -1 & 1 \end{vmatrix} = (-1)(1) - (8)(-1) = -1 + 8 = 7 \)

2. \( \begin{vmatrix} 4 & 8 \\ 0 & 1 \end{vmatrix} = (4)(1) - (8)(0) = 4 \)

3. \( \begin{vmatrix} 4 & -1 \\ 0 & -1 \end{vmatrix} = (4)(-1) - (-1)(0) = -4 \)

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 7\hat{i} - 4\hat{j} - 4\hat{k} \]

### Step 3: Express the required vector \( \mathbf{C} \)
Let \( \mathbf{C} = \lambda (\mathbf{A} \times \mathbf{B}) = \lambda (7\hat{i} - 4\hat{j} - 4\hat{k}) \)

### Step 4: Find the magnitude of \( \mathbf{C} \)
The magnitude of \( \mathbf{C} \) is given by:
\[ |\mathbf{C}| = |\lambda| \sqrt{7^2 + (-4)^2 + (-4)^2} = |\lambda| \sqrt{49 + 16 + 16} = |\lambda| \sqrt{81} = 9|\lambda| \]

### Step 5: Set the magnitude equal to 15
We need:
\[ 9|\lambda| = 15 \]
Solving for \( |\lambda| \):
\[ |\lambda| = \frac{15}{9} = \frac{5}{3} \]

### Step 6: Write the final vector \( \mathbf{C} \)
Thus, the vector \( \mathbf{C} \) can be expressed as:
\[ \mathbf{C} = \frac{5}{3} (7\hat{i} - 4\hat{j} - 4\hat{k}) = \frac{35}{3}\hat{i} - \frac{20}{3}\hat{j} - \frac{20}{3}\hat{k} \]

### Final Answer
The required vector \( \mathbf{C} \) is:
\[ \mathbf{C} = \frac{35}{3}\hat{i} - \frac{20}{3}\hat{j} - \frac{20}{3}\hat{k} \]