Question

If the square of the shortest distance between the lines \[ \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \] and \[ \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5} \] is \( \frac{m}{n} \), where \( m \) and \( n \) are co-prime numbers, then \( m+n \) is equal to:

• A. 6
• B. 9
• C. 21
• D. 14

Hint:

When finding the shortest distance between two skew lines, use the cross product of the direction vectors and the vector between points on the lines. Then apply the formula for the distance.

Answer 1

The Correct Option is B

To determine the shortest distance between the provided skew lines:

The lines are represented in vector form:

• Line 1: \( \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \), which can be expressed as \(\vec{r}_1 = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}\).
• Line 2: \( \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5} \), which can be expressed as \(\vec{r}_2 = \begin{pmatrix} -1 \\ -3 \\ -5 \end{pmatrix} + s\begin{pmatrix} 2 \\ 4 \\ -5 \end{pmatrix}\).

The direction vectors for these lines are:

• \(\vec{a}_1 = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}\)
• \(\vec{a}_2 = \begin{pmatrix} 2 \\ 4 \\ -5 \end{pmatrix}\)

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

\(d = \frac{|(\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|}\)

Where:

• \(\vec{b}_1 = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}\)
• \(\vec{b}_2 = \begin{pmatrix} -1 \\ -3 \\ -5 \end{pmatrix}\)

First, calculate the cross product \(\vec{a}_1 \times \vec{a}_2\):

\(\vec{a}_1 \times \vec{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(2 \times -5 - 4 \times -3) - \hat{j}(1 \times -5 - 2 \times -3) + \hat{k}(1 \times 4 - 2 \times 2)\)

\(\vec{a}_1 \times \vec{a}_2 = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4)\)

\(\vec{a}_1 \times \vec{a}_2 = 2\hat{i} - 1\hat{j} + 0\hat{k}\)

Therefore, \(\vec{a}_1 \times \vec{a}_2 = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}\).

Next, compute the magnitude of \(\vec{a}_1 \times \vec{a}_2\):

\(|\vec{a}_1 \times \vec{a}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5}\)

Then, calculate the difference \(\vec{b}_2 - \vec{b}_1\):

\(\vec{b}_2 - \vec{b}_1 = \begin{pmatrix} -1 \\ -3 \\ -5 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} -3 \\ -4 \\ -2 \end{pmatrix}\)

Now, compute the dot product \((\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)\):

\((\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2) = \begin{pmatrix} -3 \\ -4 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = (-3)(2) + (-4)(-1) + (-2)(0) = -6 + 4 + 0 = -2\)

The absolute value of the dot product is \(|-2| = 2\).

The shortest distance \(d\) is calculated as:

\(d = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\)

The square of the shortest distance is:

\(\left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{4 \times 5}{25} = \frac{20}{25} = \frac{4}{5}\)

Given that this value \(\frac{m}{n} = \frac{4}{5}\), where \(m\) and \(n\) are coprime, we have \(m = 4\) and \(n = 5\). Thus, \(m + n = 4 + 5 = 9\).