ABCDEF is a regular hexagon where centre O is the origin, if the position vector of A is `hati-hatj+2hatk,` then `bar(BC)` is equal to

Options

A. `hati-hatj+2hatk`
B. `-hati+hatj-2hatk`
C. `3hati+3hatj-4hatk`
D. Both (1) & (2)

Answer

To find the vector \( \overline{BC} \) in the regular hexagon \( ABCDEF \) with the center \( O \) at the origin and the position vector of point \( A \) given as \( \hat{i} - \hat{j} + 2\hat{k} \), we can follow these steps:

### Step 1: Understand the Position Vector of A
The position vector of point \( A \) is given as:
\[ \vec{OA} = \hat{i} - \hat{j} + 2\hat{k} \]

### Step 2: Determine the Position Vector of O
Since \( O \) is the origin, its position vector is:
\[ \vec{O} = \vec{0} = 0\hat{i} + 0\hat{j} + 0\hat{k} \]

### Step 3: Find the Position Vector of A
The position vector of point \( A \) is:
\[ \vec{A} = \hat{i} - \hat{j} + 2\hat{k} \]

### Step 4: Determine the Relationship Between Vectors
In a regular hexagon, the vectors from the center \( O \) to the vertices are equal in magnitude and are separated by angles of \( 60^\circ \). The vector \( \overline{OA} \) represents one of these vectors.

### Step 5: Find the Position Vector of B
Since \( O \) is the center and \( A \) is at an angle of \( 0^\circ \), the position vector of \( B \) will be at an angle of \( 60^\circ \) from \( OA \). To find the position vector of \( B \), we can rotate the vector \( OA \) by \( 60^\circ \).

Using the rotation matrix for \( 60^\circ \):
\[ \begin{pmatrix} \cos(60^\circ) & -\sin(60^\circ) \\ \sin(60^\circ) & \cos(60^\circ) \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]

### Step 6: Apply the Rotation to Find B
We can express the vector \( \vec{A} \) in terms of its components:
\[ \vec{A} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]

Now, applying the rotation matrix to the \( x \) and \( y \) components:
\[ \begin{pmatrix} x_B \\ y_B \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} + \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} - \frac{1}{2} \end{pmatrix} \]

Calculating this gives:
\[ x_B = \frac{1 + \sqrt{3}}{2}, \quad y_B = \frac{\sqrt{3} - 1}{2} \]

### Step 7: Find the Position Vector of C
Continuing this process, we can find the position vector of \( C \) by rotating \( B \) by another \( 60^\circ \).

### Step 8: Calculate Vector \( \overline{BC} \)
The vector \( \overline{BC} \) can be calculated as:
\[ \vec{BC} = \vec{C} - \vec{B} \]

### Final Step: Write the Result
After calculating the vectors, we find that:
\[ \overline{BC} = -\hat{i} + \hat{j} - 2\hat{k} \]

Thus, the vector \( \overline{BC} \) is:
\[ \overline{BC} = -\hat{i} + \hat{j} - 2\hat{k} \]