In a tetrahedron OABC, if `OA=hat(i), OB=hat(i)+hat(j) and OC=hat(i)+2hat(j)+hat(k)`,if shortest distance between egdes OA and BC is m, then `sqrt(2)m` is equal to …(where O is the origin).

Answer

To solve the problem, we need to find the shortest distance between the edges OA and BC in the tetrahedron OABC, where the points are defined as follows:

- \( O = (0, 0, 0) \)
- \( A = (1, 0, 0) \)
- \( B = (1, 1, 0) \)
- \( C = (1, 2, 1) \)

### Step 1: Define the Lines OA and BC

The line OA can be represented parametrically as:
\[ \text{OA: } (x, y, z) = (t, 0, 0) \quad \text{for } t \in [0, 1] \]

The line BC can be represented parametrically as:
\[ \text{BC: } (x, y, z) = (1, 1 + s, s) \quad \text{for } s \in [0, 1] \]

### Step 2: Find Direction Ratios

The direction ratios for line OA are:
\[ \text{Direction of OA: } (1, 0, 0) \]

The direction ratios for line BC are:
\[ \text{Direction of BC: } (0, 1, 1) \]

### Step 3: Use the Formula for Distance Between Skew Lines

The formula for the distance \( d \) between two skew lines given by points \( P_1 \) and \( P_2 \) and direction vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ d = \frac{|(\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a} \times \mathbf{b})|}{|\mathbf{a} \times \mathbf{b}|} \]

Where:
- \( \mathbf{P_1} = O = (0, 0, 0) \)
- \( \mathbf{P_2} = B = (1, 1, 0) \)
- \( \mathbf{a} = (1, 0, 0) \)
- \( \mathbf{b} = (0, 1, 1) \)

### Step 4: Calculate \( \mathbf{P_2} - \mathbf{P_1} \)

\[ \mathbf{P_2} - \mathbf{P_1} = (1, 1, 0) - (0, 0, 0) = (1, 1, 0) \]

### Step 5: Calculate the Cross Product \( \mathbf{a} \times \mathbf{b} \)

\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(0 \cdot 1 - 0 \cdot 1) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 1 - 0 \cdot 0) = \hat{i}(0) - \hat{j}(1) + \hat{k}(1) = (-1, 0, 1) \]

### Step 6: Calculate the Magnitude of the Cross Product

\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

### Step 7: Calculate the Dot Product

\[ (\mathbf{P_2} - \mathbf{P_1}) \cdot (\mathbf{a} \times \mathbf{b}) = (1, 1, 0) \cdot (-1, 0, 1) = 1 \cdot (-1) + 1 \cdot 0 + 0 \cdot 1 = -1 \]

### Step 8: Calculate the Distance \( d \)

\[ d = \frac{| -1 |}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 9: Relate \( d \) to \( m \)

Given that the shortest distance \( m = \frac{1}{\sqrt{2}} \), we need to find \( \sqrt{2}m \):

\[ \sqrt{2}m = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \]

### Final Answer

Thus, \( \sqrt{2}m = 1 \).

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