Shortest distance between z-axis and the line `(x-2)/(3)=(y-5)/(2)=(z+1)/(-5)` is

Options

A. `1//sqrt(13)`
B. `11//sqrt(13)`
C. `sqrt(11)//13`
D. `11//sqrt(13)`

Answer

To find the shortest distance between the z-axis and the given line represented by the equation \((x-2)/3 = (y-5)/2 = (z+1)/(-5)\), we will follow these steps:

### Step 1: Write the equations in vector form
The given line can be expressed in vector form. The equation \((x-2)/3 = (y-5)/2 = (z+1)/(-5)\) can be rewritten as:
\[ \mathbf{r} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 2 \\ -5 \end{pmatrix} \]
where \(\lambda\) is a parameter.

The z-axis can be represented as:
\[ \mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]
where \(\mu\) is another parameter.

### Step 2: Identify the direction vectors and points
From the vector forms, we identify:
- For the line:
- Point \( \mathbf{a_1} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} \)
- Direction vector \( \mathbf{b_1} = \begin{pmatrix} 3 \\ 2 \\ -5 \end{pmatrix} \)

- For the z-axis:
- Point \( \mathbf{a_2} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
- Direction vector \( \mathbf{b_2} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \)

### Step 3: Use the formula for the distance between two lines
The formula for the shortest distance \(d\) between two skew lines given by:
\[ d = \frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

### Step 4: Calculate \(\mathbf{b_1} \times \mathbf{b_2}\)
To find the cross product \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ \mathbf{b_1} = \begin{pmatrix} 3 \\ 2 \\ -5 \end{pmatrix}, \quad \mathbf{b_2} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]
Using the determinant method:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & -5 \\ 0 & 0 & 1 \end{vmatrix} = \mathbf{i}(2 \cdot 1 - 0 \cdot -5) - \mathbf{j}(3 \cdot 1 - 0 \cdot -5) + \mathbf{k}(3 \cdot 0 - 2 \cdot 0) \]
\[ = 2\mathbf{i} - 3\mathbf{j} + 0\mathbf{k} = \begin{pmatrix} 2 \\ -3 \\ 0 \end{pmatrix} \]

### Step 5: Calculate \(|\mathbf{b_1} \times \mathbf{b_2}|\)
The magnitude is calculated as:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{2^2 + (-3)^2 + 0^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 6: Calculate \(\mathbf{a_2} - \mathbf{a_1}\)
\[ \mathbf{a_2} - \mathbf{a_1} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} = \begin{pmatrix} -2 \\ -5 \\ 1 \end{pmatrix} \]

### Step 7: Calculate \((\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})\)
\[ (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) = \begin{pmatrix} 2 \\ -3 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ -5 \\ 1 \end{pmatrix} = 2 \cdot -2 + (-3) \cdot -5 + 0 \cdot 1 = -4 + 15 + 0 = 11 \]

### Step 8: Calculate the distance \(d\)
Now substituting into the distance formula:
\[ d = \frac{|11|}{\sqrt{13}} = \frac{11}{\sqrt{13}} \]

### Final Answer
The shortest distance between the z-axis and the given line is:
\[ \frac{11}{\sqrt{13}} \text{ units} \]