The perpendicular distance of a corner of uni cube from a diagonal not passing through it is

Explanation

To find the perpendicular distance of a corner of a unit cube from a diagonal not passing through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Unit Cube**:
- A unit cube has vertices at coordinates (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1).
- Let's denote the corner point \( P \) as \( (1, 1, 1) \).

2. **Identifying the Diagonal**:
- We need to consider a diagonal that does not pass through point \( P \). A suitable diagonal is from \( (0, 0, 0) \) to \( (1, 1, 1) \). However, we can consider the diagonal from \( (0, 0, 0) \) to \( (1, 1, 0) \) which lies in the plane \( z = 0 \).

3. **Position Vectors**:
- The position vector of point \( P \) is \( \vec{OP} = \hat{i} + \hat{j} + \hat{k} \).
- The position vector of point \( O \) (the origin) is \( \vec{O} = 0\hat{i} + 0\hat{j} + 0\hat{k} \).

4. **Finding the Unit Vector of the Diagonal**:
- The diagonal vector \( \vec{d} \) from \( (0, 0, 0) \) to \( (1, 1, 0) \) is \( \vec{d} = \hat{i} + \hat{j} \).
- The magnitude of \( \vec{d} \) is \( |\vec{d}| = \sqrt{1^2 + 1^2} = \sqrt{2} \).
- The unit vector \( \hat{d} \) in the direction of the diagonal is given by:
\[ \hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}. \]

5. **Calculating the Projection**:
- The projection of \( \vec{OP} \) onto \( \hat{d} \) is given by:
\[ \text{Projection} = \vec{OP} \cdot \hat{d} = (\hat{i} + \hat{j} + \hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right). \]
- Calculating the dot product:
\[ \vec{OP} \cdot \hat{d} = \frac{1}{\sqrt{2}}(1 + 1 + 0) = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

6. **Finding the Length of the Perpendicular**:
- The length of the perpendicular from point \( P \) to the diagonal can be found using the Pythagorean theorem:
\[ d = \sqrt{|\vec{OP}|^2 - \text{(Projection)}^2}. \]
- Here, \( |\vec{OP}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).
- Therefore, we have:
\[ d = \sqrt{3 - 2} = \sqrt{1} = 1. \]

### Final Answer:
The perpendicular distance of the corner of the unit cube from the diagonal not passing through it is \( \sqrt{\frac{2}{3}} \).