Find the magnitude of the shortest distance between the lines `(x)/(2)=(y)/(-3)=(z)/(1) and (x-2)/(3)=(y-1)/(-5)=(z+2)/(2)`.

Answer

To find the magnitude of the shortest distance between the two lines given by the equations

\[ \frac{x}{2} = \frac{y}{-3} = \frac{z}{1} \]

and

\[ \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}, \]

we can follow these steps:

### Step 1: Write the equations of the lines in vector form

The first line can be expressed as:

\[ \mathbf{L_1}: \mathbf{r} = \mathbf{a_1} + \lambda \mathbf{b_1} \]

where \(\mathbf{a_1} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}\) and \(\mathbf{b_1} = 2\mathbf{i} - 3\mathbf{j} + 1\mathbf{k}\).

The second line can be expressed as:

\[ \mathbf{L_2}: \mathbf{r} = \mathbf{a_2} + \mu \mathbf{b_2} \]

where \(\mathbf{a_2} = 2\mathbf{i} + 1\mathbf{j} - 2\mathbf{k}\) and \(\mathbf{b_2} = 3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}\).

### Step 2: Find the cross product of the direction vectors

We need to calculate \(\mathbf{b_1} \times \mathbf{b_2}\):

\[ \mathbf{b_1} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}, \quad \mathbf{b_2} = \begin{pmatrix} 3 \\ -5 \\ 2 \end{pmatrix} \]

Using the determinant to find the cross product:

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]

Calculating this determinant:

\[ = \mathbf{i}((-3)(2) - (1)(-5)) - \mathbf{j}((2)(2) - (1)(3)) + \mathbf{k}((2)(-5) - (-3)(3)) \]

\[ = \mathbf{i}(-6 + 5) - \mathbf{j}(4 - 3) + \mathbf{k}(-10 + 9) \]

\[ = -\mathbf{i} - \mathbf{j} - \mathbf{k} \]

Thus,

\[ \mathbf{b_1} \times \mathbf{b_2} = -\mathbf{i} - \mathbf{j} - \mathbf{k} \]

### Step 3: Find the magnitude of the cross product

The magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\) is:

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]

### Step 4: Find the vector \( \mathbf{a_2} - \mathbf{a_1} \)

Now, we find \(\mathbf{a_2} - \mathbf{a_1}\):

\[ \mathbf{a_2} - \mathbf{a_1} = (2\mathbf{i} + 1\mathbf{j} - 2\mathbf{k}) - (0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k}) = 2\mathbf{i} + 1\mathbf{j} - 2\mathbf{k} \]

### Step 5: Calculate the dot product

Now, we compute the dot product:

\[ (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) = (-\mathbf{i} - \mathbf{j} - \mathbf{k}) \cdot (2\mathbf{i} + 1\mathbf{j} - 2\mathbf{k}) \]

Calculating this:

\[ = -1 \cdot 2 - 1 \cdot 1 - 1 \cdot (-2) = -2 - 1 + 2 = -1 \]

### Step 6: Use the formula for the shortest distance

The formula for the shortest distance \(d\) between the two lines is given by:

\[ d = \frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

Substituting the values we found:

\[ d = \frac{|-1|}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer

Thus, the magnitude of the shortest distance between the given two lines is

\[ \frac{1}{\sqrt{3}} \text{ units.} \]

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