The unit vector which is orhtogonal to the vector `3hati+2hatj+6hatk` and is coplanar with vectors `2hati+hatj+hatk` and `hati-hatj+hatk`, is

Options

A. `1/(sqrt(41))(2hati-6hatj+hatk)`
B. `1/(sqrt(13))(2hati-3hatj)`
C. `1/(sqrt(10))(3hatj-hatk)`
D. `1/(sqrt(34))(4hati+3hatj-3hatk)`

Explanation

To find the unit vector that is orthogonal to the vector \( \mathbf{A} = 3\hat{i} + 2\hat{j} + 6\hat{k} \) and coplanar with 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: Define the unit vector
Let the unit vector be \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \). Since it is a unit vector, its magnitude must equal 1:
\[ \sqrt{x^2 + y^2 + z^2} = 1 \]
Squaring both sides, we get:
\[ x^2 + y^2 + z^2 = 1 \quad \text{(Equation 1)} \]

### Step 2: Use the orthogonality condition
The vector \( \mathbf{r} \) is orthogonal to \( \mathbf{A} \), which means their dot product must be zero:
\[ \mathbf{r} \cdot \mathbf{A} = 0 \]
Calculating the dot product:
\[ 3x + 2y + 6z = 0 \quad \text{(Equation 2)} \]

### Step 3: Use the coplanarity condition
The vectors \( \mathbf{B} \), \( \mathbf{C} \), and \( \mathbf{r} \) are coplanar, which means the scalar triple product must be zero. We can express this condition using the determinant:
\[ \begin{vmatrix} x & y & z \\ 2 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = 0 \]
Calculating the determinant:
\[ x(1 \cdot 1 - (-1) \cdot 1) - y(2 \cdot 1 - 1 \cdot 1) + z(2 \cdot (-1) - 1 \cdot 1) = 0 \]
This simplifies to:
\[ x(1 + 1) - y(2 - 1) + z(-2 - 1) = 0 \]
\[ 2x - y - 3z = 0 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
Now we have a system of equations:
1. \( x^2 + y^2 + z^2 = 1 \) (Equation 1)
2. \( 3x + 2y + 6z = 0 \) (Equation 2)
3. \( 2x - y - 3z = 0 \) (Equation 3)

From Equation 3, we can express \( y \) in terms of \( x \) and \( z \):
\[ y = 2x - 3z \]

### Step 5: Substitute into Equation 2
Substituting \( y \) into Equation 2:
\[ 3x + 2(2x - 3z) + 6z = 0 \]
This simplifies to:
\[ 3x + 4x - 6z + 6z = 0 \implies 7x = 0 \implies x = 0 \]

### Step 6: Substitute \( x \) into the other equations
Substituting \( x = 0 \) into Equation 3:
\[ 2(0) - y - 3z = 0 \implies -y - 3z = 0 \implies y = -3z \]

### Step 7: Substitute \( y \) into Equation 1
Substituting \( y = -3z \) into Equation 1:
\[ 0^2 + (-3z)^2 + z^2 = 1 \]
\[ 9z^2 + z^2 = 1 \implies 10z^2 = 1 \implies z^2 = \frac{1}{10} \implies z = \pm \frac{1}{\sqrt{10}} \]

### Step 8: Find \( y \)
Using \( z \) to find \( y \):
\[ y = -3z = -3\left(\pm \frac{1}{\sqrt{10}}\right) = \mp \frac{3}{\sqrt{10}} \]

### Step 9: Write the unit vector
Thus, the unit vector \( \mathbf{r} \) can be expressed as:
\[ \mathbf{r} = 0\hat{i} + \left(-\frac{3}{\sqrt{10}}\right)\hat{j} + \left(\frac{1}{\sqrt{10}}\right)\hat{k} \]
or
\[ \mathbf{r} = 0\hat{i} + \left(\frac{3}{\sqrt{10}}\right)\hat{j} + \left(-\frac{1}{\sqrt{10}}\right)\hat{k} \]

### Final Answer
The two possible unit vectors are:
\[ \mathbf{r} = -\frac{3}{\sqrt{10}}\hat{j} + \frac{1}{\sqrt{10}}\hat{k} \quad \text{or} \quad \mathbf{r} = \frac{3}{\sqrt{10}}\hat{j} - \frac{1}{\sqrt{10}}\hat{k} \]