The shortest distance between the lines `(x-4)/(4)=(y+2)/(5)=(z+3)/(3)` and `(x-1)/(3)=(y-3)/(4)=(z-4)/(2) ` is

Options

A. `6sqrt3`
B. `6sqrt2`
C. `2sqrt6`
D. `3sqrt6`

Answer

To find the shortest distance between the two given lines, we can follow these steps:

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

The first line can be expressed as:
\[ \frac{x-4}{4} = \frac{y+2}{5} = \frac{z+3}{3} \]
This can be rewritten in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix} \]
where \(\begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix}\) is a point on the line and \(\begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix}\) is the direction vector.

The second line can be expressed as:
\[ \frac{x-1}{3} = \frac{y-3}{4} = \frac{z-4}{2} \]
This can be rewritten in vector form as:
\[ \mathbf{r_2} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix} \]
where \(\begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}\) is a point on the line and \(\begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}\) is the direction vector.

### Step 2: Identify the points and direction vectors.

From the above, we have:
- For line 1:
- Point \(A_1 = \begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix}\)
- Direction vector \(B_1 = \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix}\)

- For line 2:
- Point \(A_2 = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}\)
- Direction vector \(B_2 = \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}\)

### Step 3: Calculate \(A_2 - A_1\).

\[ A_2 - A_1 = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 - 4 \\ 3 - (-2) \\ 4 - (-3) \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \\ 7 \end{pmatrix} \]

### Step 4: Compute the cross product \(B_1 \times B_2\).

\[ B_1 \times B_2 = \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix} \times \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix} \]

Using the determinant method:
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} \]

Calculating this determinant:
\[ = \mathbf{i}(5 \cdot 2 - 3 \cdot 4) - \mathbf{j}(4 \cdot 2 - 3 \cdot 3) + \mathbf{k}(4 \cdot 4 - 5 \cdot 3) \]
\[ = \mathbf{i}(10 - 12) - \mathbf{j}(8 - 9) + \mathbf{k}(16 - 15) \]
\[ = -2\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \]

### Step 5: Calculate the magnitude of the cross product.

\[ |B_1 \times B_2| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 6: Calculate the shortest distance \(D\).

The formula for the shortest distance \(D\) between two skew lines is given by:
\[ D = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} \]

Calculating the dot product:
\[ (A_2 - A_1) \cdot (B_1 \times B_2) = \begin{pmatrix} -3 \\ 5 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \]
\[ = (-3)(-2) + (5)(1) + (7)(1) = 6 + 5 + 7 = 18 \]

Now substituting back into the distance formula:
\[ D = \frac{|18|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6} \]

### Final Answer:
The shortest distance between the lines is \(3\sqrt{6}\).

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