Q14:

Find the shortest distance between the lines whose vector equations are $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$

Step 1: Identification of Parameters

The shortest distance $d$ between two skew lines given by $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is calculated using the formula: $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ [Per the formula for the shortest distance between skew lines].

From the given equations:

• $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$
• $\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$
• $\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$
• $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$

Step 2: Vector Subtraction and Cross Product

First, calculate $(\vec{a}_2 - \vec{a}_1)$:

$(\vec{a}_2 - \vec{a}_1) = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$

Next, calculate the cross product $\vec{b}_1 \times \vec{b}_2$ using the determinant method:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}$ $= \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6))$ $= -9\hat{i} + 3\hat{j} + 9\hat{k}$

Step 3: Calculating Magnitude and Dot Product

Calculate the magnitude $|\vec{b}_1 \times \vec{b}_2|$:

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + (3)^2 + (9)^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$

Calculate the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$:

$(3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = (3 \times -9) + (3 \times 3) + (3 \times 9) = -27 + 9 + 27 = 9$

Step 4: Final Distance Calculation

Substitute the values into the formula:

$d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}}$

Final Solution: The shortest distance is $\frac{3}{\sqrt{19}}$ units (or rationalized as $\frac{3\sqrt{19}}{19}$ units).