Question

Find the shortest distance between the lines \[ \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \] \[ \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \]

Hint:

For parallel lines, use $d = \dfrac{|\overrightarrow{AB}\times \vec{d}|}{|\vec{d}|}$.

Answer 1

Step 1: Write in standard form. First line passes through $A(1,2,-4)$, direction vector $\vec{d} = (2,3,6)$. Second line passes through $B(3,3,-5)$, direction vector $\vec{d} = (2,3,6)$. Step 2: Check relation of lines. Since both lines have the same direction vector $(2,3,6)$, they are parallel. Step 3: Formula for shortest distance between parallel lines. \[ d = \frac{|\overrightarrow{AB} \times \vec{d}|}{|\vec{d}|} \] Step 4: Find $\overrightarrow{AB}$. \[ \overrightarrow{AB} = (3-1,\,3-2,\,-5-(-4)) = (2,1,-1) \] Step 5: Compute cross product. \[ \overrightarrow{AB} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \] \[ = \hat{i}(1\cdot 6 - (-1)\cdot 3) - \hat{j}(2\cdot 6 - (-1)\cdot 2) + \hat{k}(2\cdot 3 - 1\cdot 2) \] \[ = \hat{i}(6+3) - \hat{j}(12+2) + \hat{k}(6-2) \] \[ = 9\hat{i} - 14\hat{j} + 4\hat{k} \] Step 6: Magnitudes. \[ |\overrightarrow{AB} \times \vec{d}| = \sqrt{9^{2} + (-14)^{2} + 4^{2}} = \sqrt{81+196+16} = \sqrt{293} \] \[ |\vec{d}| = \sqrt{2^{2}+3^{2}+6^{2}} = \sqrt{4+9+36} = \sqrt{49} = 7 \] Step 7: Distance. \[ d = \frac{\sqrt{293}}{7} \]

Final Answer: \[ \boxed{d = \dfrac{\sqrt{293}}{7}} \]