The shortest distance between the lines \(\frac{x−3}{2}=\frac{y−2}{3}=\frac{z−1}{−1}\) and \(\frac{x+3}{2}=\frac{y-6}{1}=\frac{z−5}{3}\) is

Question

Let \( P(\alpha, \beta, \gamma) \) be the point on the line \[ \frac{x-1}{2} = \frac{y+1}{-3} = z \] at a distance \( 4\sqrt{14} \) from the point \( (1,-1,0) \) and nearer to the origin. Then the shortest distance between the lines \[ \frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3} \quad \text{and} \quad \frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1} \] is equal to

Answer 1

To find the shortest distance between two skew lines, we use the formula:

d = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|}

Where:

\(\vec{r_1}\) and \(\vec{r_2}\) are position vectors of points on the first and second line respectively.
\(\vec{a_1}\) and \(\vec{a_2}\) are direction vectors of the first and second line respectively.

For the line \(\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1}\), the position vector can be \(\vec{r_1} = 3\hat{i} + 2\hat{j} + 1\hat{k}\) and the direction vector \(\vec{a_1} = 2\hat{i} + 3\hat{j} - 1\hat{k}\).

For the line \(\frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}\), the position vector can be \(\vec{r_2} = -3\hat{i} + 6\hat{j} + 5\hat{k}\) and the direction vector \(\vec{a_2} = 2\hat{i} + 1\hat{j} + 3\hat{k}\).

First, compute \(\vec{a_1} \times \vec{a_2}\):

\(\vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix}\)

This results in:

\(\vec{a_1} \times \vec{a_2} = (9 + 1)\hat{i} - (-6 - 2)\hat{j} + (2 \cdot 1 - 3 \cdot 3)\hat{k} = 10\hat{i} + 8\hat{j} - 7\hat{k}\)

Next, compute the magnitude |\vec{a_1} \times \vec{a_2}|:

|\vec{a_1} \times \vec{a_2}| = \sqrt{10^2 + 8^2 + (-7)^2} = \sqrt{100 + 64 + 49} = \sqrt{213}

Now find (\vec{r_2} - \vec{r_1}):

\vec{r_2} - \vec{r_1} = (-3 - 3)\hat{i} + (6 - 2)\hat{j} + (5 - 1)\hat{k} = -6\hat{i} + 4\hat{j} + 4\hat{k}

Calculate (\vec{r_2} - \vec{r_1}) \cdot (\vec{a_1} \times \vec{a_2}):

(-6\hat{i} + 4\hat{j} + 4\hat{k}) \cdot (10\hat{i} + 8\hat{j} - 7\hat{k}) = -60 + 32 - 28 = -56

The shortest distance is:

d = \frac{|-56|}{\sqrt{213}} = \frac{56}{\sqrt{213}}

Rationalizing it, multiply numerator and the denominator by \sqrt{5} :

d = \frac{56 \cdot \sqrt{5}}{213}

This simplifies to:

\frac{18}{\sqrt{5}}

Hence, the correct answer is the shortest distance \(\frac{18}{\sqrt{5}}\).

Answer 2