A unit vector c perpendicular to a and coplanar with a and b, `a=i+j+k, b=i+2j` is given by

Options

A. `1/sqrt(2)(i+k)`
B. `1/sqrt(2)(i-j)`
C. `1/sqrt(2)(j+k)`
D. `1/sqrt(2)(-j+k)`

Answer

To find a unit vector \( \mathbf{c} \) that is perpendicular to vector \( \mathbf{a} \) and coplanar with vectors \( \mathbf{a} \) and \( \mathbf{b} \), we can follow these steps:

### Given:
- \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \)
- \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} \)

### Step 1: Define the vector \( \mathbf{c} \)
Assume \( \mathbf{c} \) can be expressed as:
\[ \mathbf{c} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \]

### Step 2: Apply the condition for perpendicularity
Since \( \mathbf{c} \) is perpendicular to \( \mathbf{a} \), we have:
\[ \mathbf{c} \cdot \mathbf{a} = 0 \]
Calculating the dot product:
\[ (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = x + y + z = 0 \]
This gives us our first equation:
\[ x + y + z = 0 \quad \text{(1)} \]

### Step 3: Apply the condition for coplanarity
For \( \mathbf{c} \) to be coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), the scalar triple product must equal zero:
\[ \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \]
First, we need to calculate \( \mathbf{a} \times \mathbf{b} \):
\[ \mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \]
Calculating the cross product:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & 2 & 0 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}(1 \cdot 0 - 1 \cdot 2) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot 2 - 1 \cdot 1) \]
\[ = -2\mathbf{i} + \mathbf{j} + \mathbf{k} \]
Thus, \( \mathbf{a} \times \mathbf{b} = -2\mathbf{i} + \mathbf{j} + \mathbf{k} \).

Now, we need to compute \( \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \):
\[ \mathbf{c} \cdot (-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -2x + y + z = 0 \quad \text{(2)} \]

### Step 4: Solve the system of equations
We now have two equations:
1. \( x + y + z = 0 \)
2. \( -2x + y + z = 0 \)

Subtracting equation (1) from equation (2):
\[ (-2x + y + z) - (x + y + z) = 0 \implies -3x = 0 \implies x = 0 \]

Substituting \( x = 0 \) into equation (1):
\[ 0 + y + z = 0 \implies y + z = 0 \implies z = -y \]

### Step 5: Express \( \mathbf{c} \)
Thus, we can express \( \mathbf{c} \) as:
\[ \mathbf{c} = 0 \cdot \mathbf{i} + y \cdot \mathbf{j} - y \cdot \mathbf{k} = y(\mathbf{j} - \mathbf{k}) \]

### Step 6: Normalize \( \mathbf{c} \) to make it a unit vector
To make \( \mathbf{c} \) a unit vector:
\[ \|\mathbf{c}\| = 1 \implies \sqrt{0^2 + y^2 + (-y)^2} = 1 \implies \sqrt{2y^2} = 1 \implies |y| = \frac{1}{\sqrt{2}} \]
Thus, we can take \( y = \frac{1}{\sqrt{2}} \) (or \( y = -\frac{1}{\sqrt{2}} \)).

### Final Expression for \( \mathbf{c} \)
Substituting \( y \):
\[ \mathbf{c} = \frac{1}{\sqrt{2}} \mathbf{j} - \frac{1}{\sqrt{2}} \mathbf{k} = \frac{1}{\sqrt{2}} (\mathbf{j} - \mathbf{k}) \]

### Conclusion
The unit vector \( \mathbf{c} \) is:
\[ \mathbf{c} = \frac{1}{\sqrt{2}} (-\mathbf{j} + \mathbf{k}) \]