The length of edge of a cube is a, then shortest distance between the diagonal of cube and an edge skew to it is

Options

A. `asqrt(2)`
B. `a//sqrt(2)`
C. `sqrt(2)//a`
D. None

Explanation

To find the shortest distance between the diagonal of a cube and an edge skew to it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Cube and Coordinates**:
- Consider a cube with edge length \( a \). We can place the cube in a 3D coordinate system.
- Let the vertices of the cube be at the following coordinates:
- \( O(0, 0, 0) \)
- \( A(a, 0, 0) \)
- \( B(0, a, 0) \)
- \( C(0, 0, a) \)
- \( D(a, a, 0) \)
- \( E(a, 0, a) \)
- \( F(0, a, a) \)
- \( G(a, a, a) \)

2. **Identifying the Diagonal and the Edge**:
- The diagonal of the cube can be represented by the line segment from \( O(0, 0, 0) \) to \( G(a, a, a) \).
- An edge skew to this diagonal can be the edge from \( A(a, 0, 0) \) to \( D(a, a, 0) \).

3. **Finding the Midpoint of the Diagonal**:
- The midpoint \( M \) of the diagonal \( OG \) can be calculated as:
\[ M = \left( \frac{0 + a}{2}, \frac{0 + a}{2}, \frac{0 + a}{2} \right) = \left( \frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) \]

4. **Finding the Direction Vector of the Edge**:
- The edge \( AD \) can be represented by the points \( A(a, 0, 0) \) and \( D(a, a, 0) \).
- The direction vector \( \vec{AD} \) is given by:
\[ \vec{AD} = D - A = (a, a, 0) - (a, 0, 0) = (0, a, 0) \]

5. **Finding the Shortest Distance**:
- The shortest distance \( d \) from a point to a line in 3D can be calculated using the formula:
\[ d = \frac{|\vec{AP} \cdot (\vec{b} \times \vec{c})|}{|\vec{b} \times \vec{c}|} \]
where \( \vec{AP} \) is the vector from point \( A \) to point \( M \), and \( \vec{b} \) and \( \vec{c} \) are direction vectors of the line.
- Here, \( \vec{AP} = M - A = \left( \frac{a}{2} - a, \frac{a}{2} - 0, \frac{a}{2} - 0 \right) = \left( -\frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) \).
- The cross product \( \vec{b} \times \vec{c} \) where \( \vec{b} = (0, a, 0) \) and \( \vec{c} = (0, 0, a) \):
\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = (a^2) \hat{i} \]
- The magnitude of \( \vec{b} \times \vec{c} \) is \( |\vec{b} \times \vec{c}| = a^2 \).

6. **Calculating the Distance**:
- Now, we need to calculate \( |\vec{AP} \cdot (\vec{b} \times \vec{c})| \):
\[ \vec{AP} \cdot (a^2, 0, 0) = \left( -\frac{a}{2}, \frac{a}{2}, \frac{a}{2} \right) \cdot (a^2, 0, 0) = -\frac{a^3}{2} \]
- Therefore, the distance \( d \) is:
\[ d = \frac{\left| -\frac{a^3}{2} \right|}{a^2} = \frac{\frac{a^3}{2}}{a^2} = \frac{a}{2} \]

### Final Result:
The shortest distance between the diagonal of the cube and an edge skew to it is \( \frac{a}{2} \).