If `a = 2i + 3j + 6k, b=3i - 6j + 2k` then `a xx b` is a vector

Options

A. perpendicular to a only
B. perpendicular to b only
C. perpendicular to both
D. none

Answer

To solve the problem of finding the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \), we will follow these steps:

### Step 1: Define the Vectors
We have:
\[ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k} \]
\[ \mathbf{b} = 3\mathbf{i} - 6\mathbf{j} + 2\mathbf{k} \]

### Step 2: Set Up the Cross Product
The cross product \( \mathbf{a} \times \mathbf{b} \) can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2 \end{vmatrix} \]

### Step 3: Calculate the Determinant
We can calculate the determinant using the formula:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} 3 & 6 \\ -6 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 6 \\ 3 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 3 \\ 3 & -6 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \mathbf{i} \):
\[ \begin{vmatrix} 3 & 6 \\ -6 & 2 \end{vmatrix} = (3)(2) - (6)(-6) = 6 + 36 = 42 \]

2. For \( \mathbf{j} \):
\[ \begin{vmatrix} 2 & 6 \\ 3 & 2 \end{vmatrix} = (2)(2) - (6)(3) = 4 - 18 = -14 \]

3. For \( \mathbf{k} \):
\[ \begin{vmatrix} 2 & 3 \\ 3 & -6 \end{vmatrix} = (2)(-6) - (3)(3) = -12 - 9 = -21 \]

### Step 4: Combine the Results
Now substituting back into the expression for the cross product:
\[ \mathbf{a} \times \mathbf{b} = 42\mathbf{i} - (-14)\mathbf{j} - 21\mathbf{k} \]
This simplifies to:
\[ \mathbf{a} \times \mathbf{b} = 42\mathbf{i} + 14\mathbf{j} - 21\mathbf{k} \]

### Step 5: Final Result
Thus, the cross product \( \mathbf{a} \times \mathbf{b} \) is:
\[ \mathbf{a} \times \mathbf{b} = 42\mathbf{i} + 14\mathbf{j} - 21\mathbf{k} \]