The shortest distance between a diagonal of a cube, of edge-length one unit and the edge not meeting it, is

Options

A. `(1)/(sqrt(3))`
B. `(1)/(sqrt(2))`
C. `sqrt(2)`
D. none of these

Answer

To find the shortest distance between a diagonal of a cube and an edge not meeting it, we can follow these steps:

### Step 1: Define the Cube and Points
Let's consider a cube with vertices at the following coordinates:
- A(0, 0, 0)
- B(1, 0, 0)
- C(1, 1, 0)
- D(0, 1, 0)
- E(0, 0, 1)
- F(1, 0, 1)
- G(1, 1, 1)
- H(0, 1, 1)

We will take the diagonal AD and the edge BH for our calculations.

### Step 2: Identify the Vectors
The diagonal AD can be represented by the vector from A to D:
- **AD** = D - A = (0, 1, 0) - (0, 0, 0) = (0, 1, 0)

The edge BH can be represented by the vector from B to H:
- **BH** = H - B = (0, 1, 1) - (1, 0, 0) = (-1, 1, 1)

### Step 3: Find the Direction Vectors
The direction vector of AD is:
- **b1** = (0, 1, 0)

The direction vector of BH is:
- **b2** = (-1, 1, 1)

### Step 4: Calculate the Cross Product
To find the shortest distance between the two lines, we need to calculate the cross product of the direction vectors b1 and b2:
- **b1 × b2** = |i j k|
|0 1 0|
|-1 1 1|

Calculating the determinant:
- = i(1*1 - 0*1) - j(0*1 - 0*(-1)) + k(0*1 - (-1)*1)
- = i(1) - j(0) + k(1)
- = (1, 0, 1)

### Step 5: Find the Vector Between Points A and B
Now we find the vector from A to B:
- **AB** = B - A = (1, 0, 0) - (0, 0, 0) = (1, 0, 0)

### Step 6: Calculate the Dot Product
Next, we calculate the dot product of AB with the cross product found earlier:
- **AB** · (b1 × b2) = (1, 0, 0) · (1, 0, 1) = 1*1 + 0*0 + 0*1 = 1

### Step 7: Calculate the Magnitude of the Cross Product
Now, we need the magnitude of the cross product:
- |b1 × b2| = √(1^2 + 0^2 + 1^2) = √(1 + 0 + 1) = √2

### Step 8: Calculate the Shortest Distance
Finally, we can find the shortest distance using the formula:
\[ \text{Distance} = \frac{|AB \cdot (b1 \times b2)|}{|b1 \times b2|} \]
Substituting the values we found:
\[ \text{Distance} = \frac{1}{\sqrt{2}} \]

### Conclusion
Thus, the shortest distance between the diagonal AD and the edge BH is:
\[ \frac{1}{\sqrt{2}} \text{ units} \]