Unit vectors a, b, c are coplanar. A unit vector d is perpendicular to them. If
`(a times b) times (c times d)=1/6i-1/3j+1/3k`
and the angle between a and b is `30^(@)`, then c is/are

Options

A. `pm1/3(-i-2j+2k)`
B. `1/3(2i+j-k)`
C. `pm1/3(-i+2j-2k)`
D. `1/3(-2i-2j+k)`

Answer

To solve the problem, we need to find the unit vector \( \mathbf{c} \) given that the unit vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are coplanar, and a unit vector \( \mathbf{d} \) is perpendicular to them. We also know that:

\[ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = \frac{1}{6} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \]

and the angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( 30^\circ \).

### Step 1: Understand the cross product identity
Using the vector triple product identity, we have:

\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \]

In our case, we can let \( \mathbf{u} = \mathbf{a} \), \( \mathbf{v} = \mathbf{b} \), and \( \mathbf{w} = \mathbf{c} \times \mathbf{d} \).

### Step 2: Apply the identity
Applying the identity, we get:

\[ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{d}) \mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{d} \]

### Step 3: Set up the equation
We can equate this to the given vector:

\[ (\mathbf{a} \cdot \mathbf{d}) \mathbf{c} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{d} = \frac{1}{6} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \]

### Step 4: Analyze the coplanarity condition
Since \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) are coplanar, the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \). This implies that:

\[ \mathbf{a} \cdot \mathbf{c} = 0 \]

### Step 5: Substitute and simplify
Substituting \( \mathbf{a} \cdot \mathbf{c} = 0 \) into our equation gives:

\[ (\mathbf{a} \cdot \mathbf{d}) \mathbf{c} = \frac{1}{6} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \]

### Step 6: Find \( \mathbf{a} \cdot \mathbf{d} \)
Since \( \mathbf{d} \) is a unit vector and is perpendicular to \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), we can denote \( \mathbf{d} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \). The angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( 30^\circ \), thus:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(30^\circ) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Step 7: Solve for \( \mathbf{c} \)
Now, we can express \( \mathbf{c} \):

\[ \mathbf{c} = \frac{1}{\mathbf{a} \cdot \mathbf{d}} \left( \frac{1}{6} \mathbf{i} - \frac{1}{3} \mathbf{j} + \frac{1}{3} \mathbf{k} \right) \]

### Step 8: Normalize \( \mathbf{c} \)
Since \( \mathbf{c} \) is a unit vector, we need to ensure that its magnitude is 1. Calculate the magnitude of the vector on the right-hand side and normalize it.

### Final Result
After performing the calculations, we find:

\[ \mathbf{c} = \frac{1}{3} \left( \mathbf{i} - 2 \mathbf{j} + 2 \mathbf{k} \right) \]