Question

Find the shortest distance between lines: \[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}, \quad \frac{x-2}{4} = \frac{y-4}{5} = \frac{z-5}{6} \]

Hint:

To find the shortest distance between two skew lines, use the formula involving the cross product of their direction vectors.

Answer 1

The shortest distance between two skew lines can be found using the formula: \[ d = \frac{| (\overrightarrow{r_2} - \overrightarrow{r_1}) \cdot (\overrightarrow{a_1} \times \overrightarrow{a_2}) |}{|\overrightarrow{a_1} \times \overrightarrow{a_2}|} \] Where \( \overrightarrow{r_1} \) and \( \overrightarrow{r_2} \) are position vectors of any points on the lines, and \( \overrightarrow{a_1} \) and \( \overrightarrow{a_2} \) are direction vectors of the lines. Step 1: The direction vectors are: \[ \overrightarrow{a_1} = \langle 2, 3, 4 \rangle, \quad \overrightarrow{a_2} = \langle 4, 5, 6 \rangle \] Step 2: The vector between the two points is: \[ \overrightarrow{r_2} - \overrightarrow{r_1} = \langle 2-1, 4-2, 5-3 \rangle = \langle 1, 2, 2 \rangle \] Step 3: Find the cross product \( \overrightarrow{a_1} \times \overrightarrow{a_2} \): \[ \overrightarrow{a_1} \times \overrightarrow{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
4 & 5 & 6 \end{vmatrix} \] \[ = \hat{i}(3 \times 6 - 4 \times 5) - \hat{j}(2 \times 6 - 4 \times 4) + \hat{k}(2 \times 5 - 3 \times 4) \] \[ = \hat{i}(18 - 20) - \hat{j}(12 - 16) + \hat{k}(10 - 12) \] \[ = \langle -2, 4, -2 \rangle \] Step 4: Find the magnitude of the cross product: \[ |\overrightarrow{a_1} \times \overrightarrow{a_2}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} \] Step 5: Find the dot product \( (\overrightarrow{r_2} - \overrightarrow{r_1}) \cdot (\overrightarrow{a_1} \times \overrightarrow{a_2}) \): \[ (\overrightarrow{r_2} - \overrightarrow{r_1}) \cdot (\overrightarrow{a_1} \times \overrightarrow{a_2}) = \langle 1, 2, 2 \rangle \cdot \langle -2, 4, -2 \rangle \] \[ = 1(-2) + 2(4) + 2(-2) = -2 + 8 - 4 = 2 \] Step 6: Find the shortest distance: \[ d = \frac{|2|}{2\sqrt{6}} = \frac{2}{2\sqrt{6}} = \frac{1}{\sqrt{6}} \] Thus, the shortest distance between the lines is: \[ \boxed{\frac{1}{\sqrt{6}}} \]