Question

The shortest distance between the lines \(\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3} \space and \space  \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\) is: =

• A. \(\frac{197}{\sqrt{179}}\)
• B. \(\frac{197}{\sqrt{191}}\)
• C. \(\frac{197}{\sqrt{189}}\)
• D. \(\frac{197}{\sqrt{237}}\)

Answer 1

The Correct Option is D

The given lines are:

\(\frac{x + 3}{1} = \frac{y-2}{2} = \frac{z+4}{3}\) and \(\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}\).

First, convert the equations into vector form. The first line can be written as:

Line 1: \(\mathbf{r_1} = \langle -3, 2, -4 \rangle + t \langle 1, 2, 3 \rangle\).

The second line can be written as:

Line 2: \(\mathbf{r_2} = \langle -3, -7, 6 \rangle + s \langle -3, 2, 4 \rangle\).

The shortest distance between two skew lines is along the common perpendicular. The direction vector of the perpendicular is given by the cross product of the direction vectors of the two lines:

\(\mathbf{d_1} = \langle 1, 2, 3 \rangle\), \(\mathbf{d_2} = \langle -3, 2, 4 \rangle\).

Cross product is:

\(\mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ -3 & 2 & 4 \end{vmatrix} = \mathbf{i}(2*4 - 3*2) - \mathbf{j}(1*4 - 3*(-3)) + \mathbf{k}(1*2 - (-3)*2)\)

\(= \langle 2, -13, 8 \rangle\).

If \(\mathbf{a_1} = \langle -3, 2, -4 \rangle\) and \(\mathbf{a_2} = \langle -3, -7, 6 \rangle\), the vector between two points on each line is \(\mathbf{a} = \mathbf{a_2} - \mathbf{a_1} = \langle 0, -9, 10 \rangle\).

The shortest distance \(d\) is found by projecting \(\mathbf{a}\) onto \(\mathbf{n}\):

\(d = \frac{|\mathbf{a} \cdot \mathbf{n}|}{|\mathbf{n}|}\).

First calculate the dot product:

\(\mathbf{a} \cdot \mathbf{n} = 0*2 + (-9)*(-13) + 10*8 = 0 + 117 + 80 = 197\).

The magnitude of the cross product:

\(|\mathbf{n}| = \sqrt{2^2 + (-13)^2 + (8)^2} = \sqrt{4 + 169 + 64} = \sqrt{237}\).

Thus, the distance is:

\(d = \frac{197}{\sqrt{237}}\)

This matches the correct answer option: \(\frac{197}{\sqrt{237}}\).