A unit vector c perpendicular to `a=i-j` and coplanar with a and `b=i+k` is

Options

A. `1/sqrt(16)(i+j+2k)`
B. `1/sqrt(3)(i-j+k)`
C. `1/sqrt(3)(i+j-k)`
D. `1/sqrt(6)(i-j+2k)`

Answer

To find a unit vector \( \mathbf{c} \) that is perpendicular to \( \mathbf{a} = \mathbf{i} - \mathbf{j} \) and coplanar with \( \mathbf{a} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{k} \), we can follow these steps:

### Step 1: Define the vector \( \mathbf{c} \)
Let \( \mathbf{c} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \).

### Step 2: Use the perpendicular condition
Since \( \mathbf{c} \) is perpendicular to \( \mathbf{a} \), we can use the dot product:
\[ \mathbf{c} \cdot \mathbf{a} = 0 \]
Calculating the dot product:
\[ (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \cdot (\mathbf{i} - \mathbf{j}) = x - y = 0 \]
This gives us our first equation:
\[ x - y = 0 \quad \Rightarrow \quad x = y \]

### Step 3: Use the coplanarity condition
The vectors \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) are coplanar if the scalar triple product is zero:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \]
Calculating \( \mathbf{b} \times \mathbf{c} \):
\[ \mathbf{b} = \mathbf{i} + \mathbf{k} \quad \Rightarrow \quad \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ x & y & z \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 0 & 1 \\ y & z \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ x & z \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 0 \\ x & y \end{vmatrix} \]
\[ = \mathbf{i}(0 \cdot z - 1 \cdot y) - \mathbf{j}(1 \cdot z - 1 \cdot x) + \mathbf{k}(1 \cdot y - 0 \cdot x) \]
\[ = -y \mathbf{i} - (z - x) \mathbf{j} + y \mathbf{k} \]
Now, we take the dot product with \( \mathbf{a} \):
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (1)(-y) + (-1)(-(z - x)) + (0)(y) = -y + z - x \]
Setting this equal to zero gives us:
\[ -y + z - x = 0 \quad \Rightarrow \quad z = x + y \]

### Step 4: Substitute \( y \) with \( x \)
Since \( x = y \), we can substitute \( y \) in the equation:
\[ z = x + x = 2x \]

### Step 5: Express \( \mathbf{c} \) in terms of \( x \)
Now we can express \( \mathbf{c} \):
\[ \mathbf{c} = x \mathbf{i} + x \mathbf{j} + 2x \mathbf{k} = x(\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \]

### Step 6: Normalize \( \mathbf{c} \)
To make \( \mathbf{c} \) a unit vector, we need to set its magnitude equal to 1:
\[ |\mathbf{c}| = \sqrt{x^2 + x^2 + (2x)^2} = \sqrt{x^2 + x^2 + 4x^2} = \sqrt{6x^2} = \sqrt{6} |x| \]
Setting this equal to 1:
\[ \sqrt{6} |x| = 1 \quad \Rightarrow \quad |x| = \frac{1}{\sqrt{6}} \]
Thus, we can take \( x = \frac{1}{\sqrt{6}} \).

### Step 7: Final expression for \( \mathbf{c} \)
Substituting \( x \) back into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = \frac{1}{\sqrt{6}} \left( \mathbf{i} + \mathbf{j} + 2\mathbf{k} \right) \]

### Final Answer
The unit vector \( \mathbf{c} \) is:
\[ \mathbf{c} = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{1}{\sqrt{6}} \mathbf{j} + \frac{2}{\sqrt{6}} \mathbf{k} \]