Question

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

Hint:

For shortest distance between skew lines, always use the formula involving scalar triple product: $\dfrac{|(\vec{AB},\vec{d}_1,\vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$.

Answer 1

Step 1: Identify points and direction vectors. - Point on line 1: $A(1,2,3)$ - Direction vector of line 1: $\vec{d}_1 = (1,-3,2)$ - Point on line 2: $B(4,5,6)$ - Direction vector of line 2: $\vec{d}_2 = (2,3,1)$ Step 2: Formula for shortest distance. \[ \text{Shortest distance} = \frac{|(\vec{AB}\cdot (\vec{d}_1 \times \vec{d}_2))|}{|\vec{d}_1 \times \vec{d}_2|} \] where $\vec{AB} = \vec{B}-\vec{A} = (3,3,3)$. Step 3: Compute cross product. \[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \] \[ = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1)-(2)(2)) + \hat{k}((1)(3)-(-3)(2)) \] \[ = \hat{i}(-3-6) - \hat{j}(1-4) + \hat{k}(3+6) \] \[ = -9\hat{i} + 3\hat{j} + 9\hat{k} \] So, $\vec{d}_1 \times \vec{d}_2 = (-9,3,9)$. Step 4: Compute numerator. \[ \vec{AB}\cdot (\vec{d}_1 \times \vec{d}_2) = (3,3,3)\cdot (-9,3,9) \] \[ = (3)(-9) + (3)(3) + (3)(9) = -27+9+27=9 \] So, $|\vec{AB}\cdot (\vec{d}_1 \times \vec{d}_2)| = 9$. Step 5: Compute denominator. \[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{(-9)^2+3^2+9^2} = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19} \] Step 6: Final result. \[ \text{Shortest distance} = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}} \]

Final Answer: \[ \boxed{\dfrac{3}{\sqrt{19}}} \]