A unit vector is orthogonal to `5hati+2hatj+6hatk` and is coplanar to `2hati+hatj+hatk and hati-hatj+hatk` then the vector is

Options

A. `(3hatj-hatk)/sqrt10`
B. `(2hati+5hatj)/sqrt29`
C. `(6hati-5hatk)/sqrt61`
D. `(2hati+2hatj-hatk)/3`

Answer

To find a unit vector that is orthogonal to the vector \( \mathbf{A} = 5\hat{i} + 2\hat{j} + 6\hat{k} \) and coplanar to the vectors \( \mathbf{B} = 2\hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{C} = \hat{i} - \hat{j} + \hat{k} \), we can follow these steps:

### Step 1: Find the cross product of vectors \( \mathbf{B} \) and \( \mathbf{C} \)

The cross product \( \mathbf{B} \times \mathbf{C} \) can be calculated using the determinant:

\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2 \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \)
3. \( \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (1)(1) = -2 - 1 = -3 \)

Putting it all together:

\[ \mathbf{B} \times \mathbf{C} = 2\hat{i} - 1\hat{j} - 3\hat{k} \]

### Step 2: Find the cross product of \( \mathbf{A} \) and \( \mathbf{B} \times \mathbf{C} \)

Now we need to find \( \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \):

\[ \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 2 & 6 \\ 2 & -1 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 2 & 6 \\ -1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 6 \\ 2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 2 \\ 2 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 6 \\ -1 & -3 \end{vmatrix} = (2)(-3) - (6)(-1) = -6 + 6 = 0 \)
2. \( \begin{vmatrix} 5 & 6 \\ 2 & -3 \end{vmatrix} = (5)(-3) - (6)(2) = -15 - 12 = -27 \)
3. \( \begin{vmatrix} 5 & 2 \\ 2 & -1 \end{vmatrix} = (5)(-1) - (2)(2) = -5 - 4 = -9 \)

Putting it all together:

\[ \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = 0\hat{i} + 27\hat{j} - 9\hat{k} = 27\hat{j} - 9\hat{k} \]

### Step 3: Find the magnitude of the resulting vector

Now, we find the magnitude of \( 27\hat{j} - 9\hat{k} \):

\[ |\mathbf{A} \times (\mathbf{B} \times \mathbf{C})| = \sqrt{(0)^2 + (27)^2 + (-9)^2} = \sqrt{0 + 729 + 81} = \sqrt{810} = 9\sqrt{10} \]

### Step 4: Find the unit vector

Finally, the unit vector \( \mathbf{\alpha} \) is given by:

\[ \mathbf{\alpha} = \frac{\mathbf{A} \times (\mathbf{B} \times \mathbf{C})}{|\mathbf{A} \times (\mathbf{B} \times \mathbf{C})|} = \frac{27\hat{j} - 9\hat{k}}{9\sqrt{10}} = \frac{3\hat{j} - \hat{k}}{\sqrt{10}} \]

### Final Answer

The unit vector is:

\[ \mathbf{\alpha} = \frac{3\hat{j} - \hat{k}}{\sqrt{10}} \]