1. Key Concepts and Formulas

• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Between Two Skew Lines: Two lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$ are skew if they are neither parallel (i.e., $\vec{b_1}$ is not parallel to $\vec{b_2}$) nor intersecting. The shortest distance, $d$, between them is given by the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ This formula represents the projection of the vector connecting a point on $L_1$ to a point on $L_2$ (i.e., $\vec{a_2} - \vec{a_1}$) onto the common perpendicular direction of the two lines (i.e., $\vec{b_1} \times \vec{b_2}$).

2. Step-by-Step Solution

Step 1: Convert Given Lines to Vector Form

We begin by expressing both lines in the standard vector form $\vec{r} = \vec{a} + \lambda \vec{b}$, which requires identifying a point on each line ($\vec{a}$) and its direction vector ($\vec{b}$).

• For Line $L_1$: The line passes through points $P_1(1, 2, 3)$ and $P_2(2, 3, 4)$.

  • Identifying a point on $L_1$ ($\vec{a_1}$): We can choose $P_1$. So, $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.
  • Identifying the direction vector $\vec{b_1}$: The direction vector is the vector connecting $P_1$ to $P_2$. $\vec{b_1} = \vec{P_1P_2} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$. Thus, $L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (\hat{i} + \hat{j} + \hat{k})$.

• For Line $L_2$: The line is given by the Cartesian equation: $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 2}{0}$.

  • Identifying a point on $L_2$ ($\vec{a_2}$): From the Cartesian form $\frac{x-x_0}{l} = \frac{y-y_0}{m} = \frac{z-z_0}{n}$, the point is $(x_0, y_0, z_0)$. Here, $x_0=1$, $y_0=-1$ (from $y+1 = y-(-1)$), and $z_0=2$. So, $\vec{a_2} = \hat{i} - \hat{j} + 2\hat{k}$.
  • Identifying the direction vector $\vec{b_2}$: The direction ratios are $(l, m, n)$. Here, $l=2$, $m=-1$, and $n=0$. So, $\vec{b_2} = 2\hat{i} - \hat{j} + 0\hat{k} = 2\hat{i} - \hat{j}$. Thus, $L_2: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu (2\hat{i} - \hat{j})$.

Step 2: Calculate Vector Connecting Points, $(\vec{a_2} - \vec{a_1})$

This vector connects a point on $L_1$ to a point on $L_2$. $\vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$ $= (1-1)\hat{i} + (-1-2)\hat{j} + (2-3)\hat{k}$ $= 0\hat{i} - 3\hat{j} - \hat{k}$

Step 3: Calculate Cross Product of Direction Vectors, $(\vec{b_1} \times \vec{b_2})$

This vector is perpendicular to both direction vectors and indicates the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix}$ $= \hat{i}((1)(0) - (1)(-1)) - \hat{j}((1)(0) - (1)(2)) + \hat{k}((1)(-1) - (1)(2))$ $= \hat{i}(0 + 1) - \hat{j}(0 - 2) + \hat{k}(-1 - 2)$ $= \hat{i} + 2\hat{j} - 3\hat{k}$

Step 4: Calculate Magnitude of Cross Product, $|\vec{b_1} \times \vec{b_2}|$

This magnitude forms the denominator of the shortest distance formula. $|\vec{b_1} \times \vec{b_2}| = |\hat{i} + 2\hat{j} - 3\hat{k}| = \sqrt{1^2 + 2^2 + (-3)^2}$ $= \sqrt{1 + 4 + 9} = \sqrt{14}$ Since $|\vec{b_1} \times \vec{b_2}| \neq 0$, the lines are not parallel, confirming they are skew lines.

Step 5: Calculate Scalar Triple Product, $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$

This dot product forms the numerator (inside the absolute value) of the shortest distance formula. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0\hat{i} - 3\hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k})$ $= (0)(1) + (-3)(2) + (-1)(-3)$ $= 0 - 6 + 3 = -3$

Step 6: Apply Shortest Distance Formula to Find $\alpha$

Now we substitute the calculated values into the shortest distance formula. $\alpha = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| = \left| \frac{-3}{\sqrt{14}} \right|$ $\alpha = \frac{|-3|}{|\sqrt{14}|} = \frac{3}{\sqrt{14}}$

Step 7: Calculate the Final Expression $28\alpha^2$

The problem asks for the value of $28\alpha^2$. We have $\alpha = \frac{3}{\sqrt{14}}$. First, calculate $\alpha^2$: $\alpha^2 = \left(\frac{3}{\sqrt{14}}\right)^2 = \frac{3^2}{(\sqrt{14})^2} = \frac{9}{14}$ Now, substitute this into the expression $28\alpha^2$: $28\alpha^2 = 28 \times \left(\frac{9}{14}\right)$ $28\alpha^2 = \frac{28}{14} \times 9$ $28\alpha^2 = 2 \times 9$ $28\alpha^2 = 18$

3. Common Mistakes & Tips

• Sign Errors: Be meticulous with signs when identifying coordinates from Cartesian forms (e.g., $y+1$ implies $y_0 = -1$) and when performing vector operations like cross products and dot products.
• Zero Denominators: A zero in the denominator of a Cartesian equation (e.g., $\frac{z-2}{0}$) simply means the direction ratio for that axis is zero, implying the line is parallel to the corresponding coordinate plane. It does not mean the term is undefined in the context of direction ratios.
• Checking for Parallelism: Before applying the skew line formula, it's good practice to quickly check if the direction vectors $\vec{b_1}$ and $\vec{b_2}$ are parallel (i.e., if one is a scalar multiple of the other). If they are, a different formula for parallel lines would be used. In this case, $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - \hat{j}$ are clearly not parallel.

4. Summary

To find the shortest distance between the two given lines, we first converted both lines into their vector forms. We then systematically calculated the necessary components: the vector connecting a point on each line $(\vec{a_2} - \vec{a_1})$, the cross product of their direction vectors $(\vec{b_1} \times \vec{b_2})$, its magnitude, and finally the scalar triple product. Substituting these values into the shortest distance formula, we found $\alpha = \frac{3}{\sqrt{14}}$. Squaring this value and multiplying by 28 yielded $28\alpha^2 = 18$.

5. Final Answer The final answer is $\boxed{1}$.