The unit vector which is orthogonal to the vector `5i+2j+6k` and is coplanar with the vectors `2i+j+k" and "i-j+k` is

Options

A. `1/sqrt(41)(2i-6j+k)`
B. `1/sqrt(29)(2i-5j)`
C. `1/sqrt(10)(3j-k)`
D. `1/sqrt(69)(2i-8j+k)`

Answer

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

### Step 1: Set up the orthogonality condition
A vector \( \mathbf{P} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) is orthogonal to \( \mathbf{A} \) if their dot product is zero:
\[ \mathbf{P} \cdot \mathbf{A} = 0 \]
This gives us the equation:
\[ 5x + 2y + 6z = 0 \tag{1} \]

### Step 2: Set up the coplanarity condition
The vector \( \mathbf{P} \) is coplanar with \( \mathbf{B} \) and \( \mathbf{C} \) if the scalar triple product is zero:
\[ \mathbf{P} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \]
First, we need to calculate \( \mathbf{B} \times \mathbf{C} \):
\[ \mathbf{B} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{C} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \]
Using the determinant:
\[ \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \]
\[ = \mathbf{i} (1 - (-1)) - \mathbf{j} (2 - 1) + \mathbf{k} (-2 - 1) \]
\[ = 2\mathbf{i} - \mathbf{j} - 3\mathbf{k} \]
Thus, \( \mathbf{B} \times \mathbf{C} = 2\mathbf{i} - \mathbf{j} - 3\mathbf{k} \).

Now, we set up the equation for the coplanarity condition:
\[ \mathbf{P} \cdot (2\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 0 \]
This gives us:
\[ 2x - y - 3z = 0 \tag{2} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 5x + 2y + 6z = 0 \) (1)
2. \( 2x - y - 3z = 0 \) (2)

From equation (2), we can express \( y \) in terms of \( x \) and \( z \):
\[ y = 2x - 3z \]
Substituting this into equation (1):
\[ 5x + 2(2x - 3z) + 6z = 0 \]
\[ 5x + 4x - 6z + 6z = 0 \]
\[ 9x = 0 \implies x = 0 \]

Substituting \( x = 0 \) back into the expression for \( y \):
\[ y = 2(0) - 3z = -3z \]

Thus, we have:
\[ \mathbf{P} = 0\mathbf{i} - 3z\mathbf{j} + z\mathbf{k} = -3z\mathbf{j} + z\mathbf{k} \]

### Step 4: Normalize the vector to find the unit vector
To find the unit vector, we need to set the magnitude of \( \mathbf{P} \) equal to 1:
\[ \|\mathbf{P}\| = \sqrt{(-3z)^2 + z^2} = \sqrt{9z^2 + z^2} = \sqrt{10z^2} = |z|\sqrt{10} \]
Setting this equal to 1:
\[ |z|\sqrt{10} = 1 \implies |z| = \frac{1}{\sqrt{10}} \implies z = \pm \frac{1}{\sqrt{10}} \]

### Step 5: Find the unit vector
Substituting \( z = \frac{1}{\sqrt{10}} \) into \( \mathbf{P} \):
\[ \mathbf{P} = -3\left(\frac{1}{\sqrt{10}}\right)\mathbf{j} + \left(\frac{1}{\sqrt{10}}\right)\mathbf{k} = -\frac{3}{\sqrt{10}}\mathbf{j} + \frac{1}{\sqrt{10}}\mathbf{k} \]
Thus, the unit vector is:
\[ \mathbf{P} = \frac{-3\mathbf{j} + \mathbf{k}}{\sqrt{10}} \]

### Final Answer
The required unit vector is:
\[ \frac{-3\mathbf{j} + \mathbf{k}}{\sqrt{10}} \]