The unit vector which is orthogonal to `a =3i+2j+6k` and coplanar with `b = 2i+j+k and c =i-j+k` is

Options

A. `(6i-5k)/(sqrt(61))`
B. `(3j-k)/(sqrt(10))`
C. `(2i-5j)/(sqrt(29))`
D. `(2i+j-2k)/(3)`

Answer

To find the unit vector that is orthogonal to the vector \( \mathbf{a} = 3\mathbf{i} + 2\mathbf{j} + 6\mathbf{k} \) 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: Define the coplanar vector
Let the required vector be \( \mathbf{r} \). Since \( \mathbf{r} \) is coplanar with \( \mathbf{b} \) and \( \mathbf{c} \), we can express \( \mathbf{r} \) as a linear combination of \( \mathbf{b} \) and \( \mathbf{c} \):
\[ \mathbf{r} = \lambda \mathbf{b} + \mu \mathbf{c} \]
Substituting the values of \( \mathbf{b} \) and \( \mathbf{c} \):
\[ \mathbf{r} = \lambda (2\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mu (\mathbf{i} - \mathbf{j} + \mathbf{k}) \]
\[ \mathbf{r} = (2\lambda + \mu)\mathbf{i} + (\lambda - \mu)\mathbf{j} + (\lambda + \mu)\mathbf{k} \]

### Step 2: Set up the orthogonality condition
Since \( \mathbf{r} \) is orthogonal to \( \mathbf{a} \), their dot product must equal zero:
\[ \mathbf{a} \cdot \mathbf{r} = 0 \]
Calculating the dot product:
\[ (3\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) \cdot ((2\lambda + \mu)\mathbf{i} + (\lambda - \mu)\mathbf{j} + (\lambda + \mu)\mathbf{k}) = 0 \]
This expands to:
\[ 3(2\lambda + \mu) + 2(\lambda - \mu) + 6(\lambda + \mu) = 0 \]
Simplifying this:
\[ 6\lambda + 3\mu + 2\lambda - 2\mu + 6\lambda + 6\mu = 0 \]
Combining like terms:
\[ (6\lambda + 2\lambda + 6\lambda) + (3\mu - 2\mu + 6\mu) = 0 \]
\[ 14\lambda + 7\mu = 0 \]

### Step 3: Solve for \( \mu \) in terms of \( \lambda \)
From the equation \( 14\lambda + 7\mu = 0 \), we can express \( \mu \) as:
\[ \mu = -2\lambda \]

### Step 4: Substitute \( \mu \) back into \( \mathbf{r} \)
Substituting \( \mu = -2\lambda \) into the expression for \( \mathbf{r} \):
\[ \mathbf{r} = (2\lambda - 2\lambda)\mathbf{i} + (\lambda + 2\lambda)\mathbf{j} + (\lambda - 2\lambda)\mathbf{k} \]
This simplifies to:
\[ \mathbf{r} = 0\mathbf{i} + 3\lambda\mathbf{j} - \lambda\mathbf{k} \]
Thus,
\[ \mathbf{r} = \lambda(0\mathbf{i} + 3\mathbf{j} - \mathbf{k}) \]

### Step 5: Find the unit vector
To find the unit vector, we need to compute the magnitude of \( \mathbf{r} \):
\[ |\mathbf{r}| = \sqrt{(0)^2 + (3\lambda)^2 + (-\lambda)^2} = \sqrt{9\lambda^2 + \lambda^2} = \sqrt{10\lambda^2} = \sqrt{10}|\lambda| \]
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{r} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{\lambda(0\mathbf{i} + 3\mathbf{j} - \mathbf{k})}{\sqrt{10}|\lambda|} = \frac{1}{\sqrt{10}}(0\mathbf{i} + 3\mathbf{j} - \mathbf{k}) \]
Thus, the unit vector is:
\[ \mathbf{u} = \frac{3}{\sqrt{10}}\mathbf{j} - \frac{1}{\sqrt{10}}\mathbf{k} \]

### Final Answer:
The unit vector which is orthogonal to \( \mathbf{a} \) and coplanar with \( \mathbf{b} \) and \( \mathbf{c} \) is:
\[ \mathbf{u} = \frac{3}{\sqrt{10}}\mathbf{j} - \frac{1}{\sqrt{10}}\mathbf{k} \]