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

Options

A. `(1)/(2)`
B. `(1)/(sqrt(2))`
C. `(1)/(sqrt(3))`
D. `(1)/(sqrt(6))`

Answer

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

### Step 1: Define the points of the cube
Let's define the vertices of the unit cube. The vertices can be represented as:
- \( O(0, 0, 0) \)
- \( A(1, 1, 1) \)
- \( B(1, 0, 1) \)
- \( C(1, 0, 0) \)
- \( D(0, 1, 0) \)
- \( E(0, 0, 1) \)
- \( F(0, 1, 1) \)
- \( G(1, 1, 0) \)

### Step 2: Identify the diagonal and the skew edge
The diagonal we will consider is \( OA \) which connects points \( O(0, 0, 0) \) and \( A(1, 1, 1) \). The skew edge we will consider is \( BC \) which connects points \( B(1, 0, 1) \) and \( C(1, 0, 0) \).

### Step 3: Find the midpoint of the diagonal
To find the midpoint \( D \) of diagonal \( OA \):
\[ D = \left( \frac{0 + 1}{2}, \frac{0 + 1}{2}, \frac{0 + 1}{2} \right) = \left( \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right) \]

### Step 4: Find the midpoint of the skew edge
To find the midpoint \( E \) of edge \( BC \):
\[ E = \left( \frac{1 + 1}{2}, \frac{0 + 0}{2}, \frac{1 + 0}{2} \right) = \left( 1, 0, \frac{1}{2} \right) \]

### Step 5: Determine the direction vectors
The direction vector of diagonal \( OA \) is:
\[ \vec{d_1} = A - O = (1, 1, 1) - (0, 0, 0) = (1, 1, 1) \]
The direction vector of edge \( BC \) is:
\[ \vec{d_2} = C - B = (1, 0, 0) - (1, 0, 1) = (0, 0, -1) \]

### Step 6: Calculate the shortest distance
The shortest distance \( D \) between the two skew lines can be calculated using the formula:
\[ D = \frac{|(\vec{d_1} \times \vec{d_2}) \cdot \vec{PQ}|}{|\vec{d_1} \times \vec{d_2}|} \]
Where \( \vec{PQ} \) is the vector connecting points \( O \) and \( B \):
\[ \vec{PQ} = B - O = (1, 0, 1) - (0, 0, 0) = (1, 0, 1) \]

### Step 7: Calculate the cross product
First, we calculate the cross product \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 0 & -1 \end{vmatrix} = \hat{i}(1 \cdot (-1) - 1 \cdot 0) - \hat{j}(1 \cdot (-1) - 1 \cdot 0) + \hat{k}(1 \cdot 0 - 1 \cdot 0) \]
\[ = -\hat{i} + \hat{j} + 0\hat{k} = (-1, 1, 0) \]

### Step 8: Calculate the magnitude of the cross product
Now, we calculate the magnitude of \( \vec{d_1} \times \vec{d_2} \):
\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 9: Calculate the dot product
Next, we calculate the dot product \( (\vec{d_1} \times \vec{d_2}) \cdot \vec{PQ} \):
\[ (-1, 1, 0) \cdot (1, 0, 1) = -1 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 = -1 \]

### Step 10: Substitute into the distance formula
Now, substitute into the distance formula:
\[ D = \frac{|(-1)|}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Conclusion
Thus, the shortest distance between the diagonal of the unit cube and the edge skew to it is:
\[ \frac{\sqrt{2}}{2} \]

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