If `a = i+j-k,b=i-j+k` and c is aunit vector perpendicular to the vector a and coplanar with a and b, then a unit vector d perpendicular to both a and c is

Options

A. `(1)/(sqrt(6)) (2i -j+k)`
B. `(1)/(sqrt2) (j+k)`
C. `(1)/(sqrt2) (i+j)`
D. `(1)/(sqrt2) (i+k)`

Answer

To solve the problem step by step, we will follow the given conditions and perform vector operations accordingly.

### Step 1: Define the vectors
Given:
- \( \mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k} \)
- \( \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k} \)

### Step 2: Find a unit vector \( \mathbf{c} \) that is perpendicular to \( \mathbf{a} \)
For \( \mathbf{c} \) to be perpendicular to \( \mathbf{a} \), we must have:
\[ \mathbf{a} \cdot \mathbf{c} = 0 \]
Let \( \mathbf{c} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \). Then:
\[ \mathbf{a} \cdot \mathbf{c} = (1)(x) + (1)(y) + (-1)(z) = x + y - z = 0 \quad \text{(Equation 1)} \]

### Step 3: Ensure \( \mathbf{c} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \)
Since \( \mathbf{c} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), we can express \( \mathbf{c} \) as:
\[ \mathbf{c} = \mathbf{a} + \lambda \mathbf{b} \]
Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{c} = (\mathbf{i} + \mathbf{j} - \mathbf{k}) + \lambda (\mathbf{i} - \mathbf{j} + \mathbf{k}) \]
This simplifies to:
\[ \mathbf{c} = (1 + \lambda)\mathbf{i} + (1 - \lambda)\mathbf{j} + (-1 + \lambda)\mathbf{k} \]

### Step 4: Substitute \( \mathbf{c} \) into Equation 1
Substituting \( \mathbf{c} \) into \( x + y - z = 0 \):
\[ (1 + \lambda) + (1 - \lambda) - (-1 + \lambda) = 0 \]
Simplifying this:
\[ 1 + \lambda + 1 - \lambda + 1 - \lambda = 0 \]
\[ 3 - \lambda = 0 \implies \lambda = 3 \]

### Step 5: Find \( \mathbf{c} \)
Substituting \( \lambda = 3 \) back into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = (1 + 3)\mathbf{i} + (1 - 3)\mathbf{j} + (-1 + 3)\mathbf{k} \]
\[ \mathbf{c} = 4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \]

### Step 6: Normalize \( \mathbf{c} \) to make it a unit vector
The magnitude of \( \mathbf{c} \) is:
\[ |\mathbf{c}| = \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \]
Thus, the unit vector \( \mathbf{c} \) is:
\[ \mathbf{c} = \frac{1}{2\sqrt{6}}(4\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) = \frac{2}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} + \frac{1}{\sqrt{6}}\mathbf{k} \]

### Step 7: Find the vector \( \mathbf{d} \) which is perpendicular to both \( \mathbf{a} \) and \( \mathbf{c} \)
To find \( \mathbf{d} \), we compute the cross product \( \mathbf{a} \times \mathbf{c} \):
\[ \mathbf{a} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 4 & -2 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{d} = \mathbf{i} \begin{vmatrix} 1 & -1 \\ -2 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -1 \\ 4 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 4 & -2 \end{vmatrix} \]
Calculating each determinant:
- For \( \mathbf{i} \): \( (1)(2) - (-1)(-2) = 2 - 2 = 0 \)
- For \( \mathbf{j} \): \( (1)(2) - (-1)(4) = 2 + 4 = 6 \)
- For \( \mathbf{k} \): \( (1)(-2) - (1)(4) = -2 - 4 = -6 \)

Thus:
\[ \mathbf{d} = 0\mathbf{i} - 6\mathbf{j} - 6\mathbf{k} = -6\mathbf{j} - 6\mathbf{k} \]

### Step 8: Normalize \( \mathbf{d} \) to make it a unit vector
The magnitude of \( \mathbf{d} \) is:
\[ |\mathbf{d}| = \sqrt{0^2 + (-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]
Thus, the unit vector \( \mathbf{d} \) is:
\[ \mathbf{d} = \frac{1}{6\sqrt{2}}(-6\mathbf{j} - 6\mathbf{k}) = -\frac{1}{\sqrt{2}}\mathbf{j} - \frac{1}{\sqrt{2}}\mathbf{k} \]

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