Key Concepts and Formulas

• Shortest Distance Between Two Skew Lines: For two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$, the shortest distance ($d$) is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ Here, $\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on the lines, and $\vec{b_1}$ and $\vec{b_2}$ are their direction vectors.
• Line Equation Forms:
  • Cartesian form: $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ corresponds to $\vec{a_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b_1} = l\hat{i} + m\hat{j} + n\hat{k}$.
  • Vector form: $\vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda (l\hat{i} + m\hat{j} + n\hat{k})$.
• Latus Rectum of an Ellipse: For an ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$:
  • If $A^2 > B^2$ (major axis along x-axis), the length of the latus rectum is $\frac{2B^2}{A}$.
  • If $B^2 > A^2$ (major axis along y-axis), the length of the latus rectum is $\frac{2A^2}{B}$.

Step-by-Step Solution

Step 1: Identify Position and Direction Vectors for Line 1 ($L_1$) The first line is given in Cartesian form: $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$. We need to convert this to the vector form $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$.

• Point on $L_1$ ($\vec{a_1}$): From the numerators $(x-(-1)), (y-0), (z-0)$, we identify the point as $(-1, 0, 0)$. So, $\vec{a_1} = -\hat{i} + 0\hat{j} + 0\hat{k} = -\hat{i}$.
• Direction vector of $L_1$ ($\vec{b_1}$): The denominators represent the direction ratios. So, $\vec{b_1} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

Step 2: Identify Position and Direction Vectors for Line 2 ($L_2$) The second line is given in vector form: $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$. This is already in the standard form $\vec{r} = \vec{a_2} + \lambda \vec{b_2}$.

• Point on $L_2$ ($\vec{a_2}$): The position vector of the point on the line is $\mathrm{p} \hat{i}+2 \hat{j}+\hat{k}$. So, $\vec{a_2} = p\hat{i} + 2\hat{j} + \hat{k}$.
• Direction vector of $L_2$ ($\vec{b_2}$): The vector multiplied by $\lambda$ is the direction vector. So, $\vec{b_2} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

We observe that $\vec{b_1}$ and $\vec{b_2}$ are not proportional ($3/2 \neq 4/3 \neq 5/4$), which confirms the lines are not parallel and are therefore skew.

Step 3: Calculate the vector $(\vec{a_2} - \vec{a_1})$ This vector connects a point on $L_1$ to a point on $L_2$, which is essential for determining their relative position in space. $\vec{a_2} - \vec{a_1} = (p\hat{i} + 2\hat{j} + \hat{k}) - (-\hat{i}) = (p+1)\hat{i} + 2\hat{j} + \hat{k}$

Step 4: Calculate the cross product $(\vec{b_1} \times \vec{b_2})$ The cross product of the direction vectors gives a vector perpendicular to both lines, indicating the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix}$ $= \hat{i}(4 \times 4 - 5 \times 3) - \hat{j}(3 \times 4 - 5 \times 2) + \hat{k}(3 \times 3 - 4 \times 2)$ $= \hat{i}(16 - 15) - \hat{j}(12 - 10) + \hat{k}(9 - 8)$ $= \hat{i} - 2\hat{j} + \hat{k}$

Step 5: Calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$ This magnitude is used in the denominator of the shortest distance formula to normalize the scalar triple product. $|\vec{b_1} \times \vec{b_2}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Step 6: Apply the Shortest Distance Formula and Solve for 'p' We are given that the shortest distance $d = \frac{1}{\sqrt{6}}$. Substitute the calculated components into the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ $\frac{1}{\sqrt{6}} = \left| \frac{((p+1)\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})}{\sqrt{6}} \right|$ First, calculate the dot product in the numerator: $((p+1)\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = (p+1)(1) + (2)(-2) + (1)(1)$ $= p+1 - 4 + 1 = p-2$ Now substitute this back into the distance equation: $\frac{1}{\sqrt{6}} = \left| \frac{p-2}{\sqrt{6}} \right|$ Multiply both sides by $\sqrt{6}$: $1 = |p-2|$ This absolute value equation gives two possibilities:

1. $p-2 = 1 \implies p = 3$
2. $p-2 = -1 \implies p = 1$ The problem states that the values of $p$ are $a, b$ with $a < b$. Therefore, $a=1$ and $b=3$.

Step 7: Formulate the Ellipse Equation The ellipse is given by $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$. Substitute the values $a=1$ and $b=3$: $\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$ $\frac{x^2}{1} + \frac{y^2}{9} = 1$

Step 8: Calculate the Length of the Latus Rectum of the Ellipse For the ellipse $\frac{x^2}{1} + \frac{y^2}{9} = 1$, we have $A^2 = 1$ and $B^2 = 9$. Since $B^2 > A^2$ (i.e., $9 > 1$), the major axis of this ellipse lies along the y-axis. The semi-major axis is $B = \sqrt{9} = 3$. The semi-minor axis is $A = \sqrt{1} = 1$. The formula for the length of the latus rectum when the major axis is along the y-axis is $\frac{2A^2}{B}$. Length of Latus Rectum $= \frac{2(1)^2}{3} = \frac{2 \times 1}{3} = \frac{2}{3}$.

Common Mistakes & Tips

• Absolute Value: Always remember the absolute value in the shortest distance formula. Forgetting it would yield only one value for p.
• Cartesian to Vector Form: Be careful when extracting coordinates from expressions like $(x+1)$. This means $x - (-1)$, so the coordinate is $-1$.
• Ellipse Latus Rectum Formula: Correctly identify the semi-major and semi-minor axes based on which denominator is larger. If the larger denominator is under $y^2$, the major axis is along the y-axis, and the formula is $\frac{2(\text{smaller denominator})}{(\text{square root of larger denominator})}$.

Summary

This problem required a two-part approach. First, we applied the shortest distance formula for skew lines to find the possible values of the parameter 'p'. This involved correctly extracting vectors from both Cartesian and vector forms of line equations, performing vector cross and dot products, and solving an absolute value equation. The two values of $p$ were identified as $a=1$ and $b=3$. In the second part, these values were used to define an ellipse. We then determined the orientation of the ellipse's major axis and applied the appropriate formula to calculate the length of its latus rectum, which came out to be $\frac{2}{3}$.

The final answer is $\boxed{\frac{2}{3}}$, which corresponds to option (A).