Key Concepts and Formulas

• Vector Equation of a Line: A line passing through a point $\vec{a}$ and parallel to a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + t\vec{b}$, where $t$ is a scalar parameter.
• Line of Shortest Distance (LSD) between Skew Lines: For two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$, the line of shortest distance is the unique line segment that is perpendicular to both $L_1$ and $L_2$. Its direction vector $\vec{d}$ is parallel to the cross product of the direction vectors of the two lines, i.e., $\vec{d} = \vec{b_1} \times \vec{b_2}$.
• Perpendicularity Condition: Two vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
• Cartesian Equation of a Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be written as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.

Step-by-Step Solution

Step 1: Convert the given Cartesian equations of the lines to their vector forms.

• Why: The vector form simplifies the identification of points on the lines and their direction vectors, which are essential for subsequent calculations involving shortest distance.

The given lines are: $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$

From these Cartesian equations, we extract the following information for each line: For $L_1$:

• A point on the line: $\vec{a_1} = (1, 2, 3)$
• Its direction vector: $\vec{b_1} = (2, 3, 4)$ The vector equation for $L_1$ is: $\vec{r} = (1\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + \lambda (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})$

For $L_2$:

• A point on the line: $\vec{a_2} = (2, 4, 5)$
• Its direction vector: $\vec{b_2} = (3, 4, 5)$ The vector equation for $L_2$ is: $\vec{r} = (2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) + \mu (3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k})$

Step 2: Determine the direction vector of the Line of Shortest Distance (LSD).

• Why: The line of shortest distance (also known as the common perpendicular) between two skew lines is perpendicular to both lines. The cross product of their direction vectors yields a vector that is perpendicular to both, thus giving the direction of the LSD. Let $\vec{d}$ be the direction vector of the LSD. It is parallel to $\vec{b_1} \times \vec{b_2}$. $\vec{d} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$ Calculating the determinant: $\vec{d} = \mathbf{i}(3 \cdot 5 - 4 \cdot 4) - \mathbf{j}(2 \cdot 5 - 4 \cdot 3) + \mathbf{k}(2 \cdot 4 - 3 \cdot 3)$ $\vec{d} = \mathbf{i}(15 - 16) - \mathbf{j}(10 - 12) + \mathbf{k}(8 - 9)$ $\vec{d} = -\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ So, the direction vector of the LSD is $(-1, 2, -1)$.

Step 3: Define general points on $L_1$ and $L_2$.

• Why: To find the specific points where the LSD intersects $L_1$ and $L_2$, we first express any arbitrary point on each line using their respective parameters $\lambda$ and $\mu$. Let $P$ be a general point on $L_1$ and $Q$ be a general point on $L_2$. Using the vector equations from Step 1: $P = (1 + 2\lambda, 2 + 3\lambda, 3 + 4\lambda)$ $Q = (2 + 3\mu, 4 + 4\mu, 5 + 5\mu)$

Step 4: Form the vector $\vec{PQ}$ and apply perpendicularity conditions.

• Why: The vector $\vec{PQ}$ connecting the points $P$ and $Q$ that define the shortest distance must itself be perpendicular to both direction vectors $\vec{b_1}$ and $\vec{b_2}$. This property allows us to set up equations to find the values of $\lambda$ and $\mu$. The vector $\vec{PQ}$ connects point $P$ on $L_1$ to point $Q$ on $L_2$: $\vec{PQ} = Q - P = ((2+3\mu) - (1+2\lambda))\mathbf{i} + ((4+4\mu) - (2+3\lambda))\mathbf{j} + ((5+5\mu) - (3+4\lambda))\mathbf{k}$ $\vec{PQ} = (1 + 3\mu - 2\lambda)\mathbf{i} + (2 + 4\mu - 3\lambda)\mathbf{j} + (2 + 5\mu - 4\lambda)\mathbf{k}$ For $\vec{PQ}$ to represent the shortest distance vector, it must be perpendicular to both $\vec{b_1}$ and $\vec{b_2}$.

• Condition 1: $\vec{PQ} \cdot \vec{b_1} = 0$ $(1 + 3\mu - 2\lambda)(2) + (2 + 4\mu - 3\lambda)(3) + (2 + 5\mu - 4\lambda)(4) = 0$ Expanding and combining like terms: $(2 + 6\mu - 4\lambda) + (6 + 12\mu - 9\lambda) + (8 + 20\mu - 16\lambda) = 0$ $(2+6+8) + (6+12+20)\mu + (-4-9-16)\lambda = 0$ $16 + 38\mu - 29\lambda = 0 \quad \Rightarrow \quad 29\lambda - 38\mu = 16 \quad \text{(Equation 1)}$

• Condition 2: $\vec{PQ} \cdot \vec{b_2} = 0$ $(1 + 3\mu - 2\lambda)(3) + (2 + 4\mu - 3\lambda)(4) + (2 + 5\mu - 4\lambda)(5) = 0$ Expanding and combining like terms: $(3 + 9\mu - 6\lambda) + (8 + 16\mu - 12\lambda) + (10 + 25\mu - 20\lambda) = 0$ $(3+8+10) + (9+16+25)\mu + (-6-12-20)\lambda = 0$ $21 + 50\mu - 38\lambda = 0 \quad \Rightarrow \quad 38\lambda - 50\mu = 21 \quad \text{(Equation 2)}$

Step 5: Solve the system of linear equations for $\lambda$ and $\mu$.

• Why: The values of $\lambda$ and $\mu$ obtained from these equations will uniquely determine the specific points $P$ and $Q$ that represent the feet of the common perpendicular on $L_1$ and $L_2$. We have the system of two linear equations:

1. $29\lambda - 38\mu = 16$
2. $38\lambda - 50\mu = 21$

To solve this system by elimination, multiply Equation 1 by 38 and Equation 2 by 29: $38 \times (29\lambda - 38\mu) = 38 \times 16 \quad \Rightarrow \quad 1102\lambda - 1444\mu = 608$ $29 \times (38\lambda - 50\mu) = 29 \times 21 \quad \Rightarrow \quad 1102\lambda - 1450\mu = 609$

Subtract the first new equation from the second new equation: $(1102\lambda - 1450\mu) - (1102\lambda - 1444\mu) = 609 - 608$ $-6\mu = 1$ $\mu = -\frac{1}{6}$

Substitute $\mu = -\frac{1}{6}$ into Equation 1: $29\lambda - 38\left(-\frac{1}{6}\right) = 16$ $29\lambda + \frac{19}{3} = 16$ $29\lambda = 16 - \frac{19}{3}$ $29\lambda = \frac{48 - 19}{3}$ $29\lambda = \frac{29}{3}$ $\lambda = \frac{1}{3}$

Step 6: Find the coordinates of point $P$ (or $Q$).

• Why: The line of shortest distance passes through $P$ and $Q$. We need the coordinates of one of these points to define the equation of the LSD. Substitute $\lambda = \frac{1}{3}$ into the coordinates of point $P$: $P = \left(1 + 2\left(\frac{1}{3}\right), 2 + 3\left(\frac{1}{3}\right), 3 + 4\left(\frac{1}{3}\right)\right)$ $P = \left(1 + \frac{2}{3}, 2 + 1, 3 + \frac{4}{3}\right)$ $P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$

Step 7: Formulate the Cartesian equation of the Line of Shortest Distance (LSD).

• Why: This equation describes the line on which the shortest distance lies. We will use it to check the given options. The LSD passes through point $P\left(\frac{5}{3}, 3, \frac{13}{3}\right)$ and has the direction vector $\vec{d} = (-1, 2, -1)$. The Cartesian equation of the LSD is: $\frac{x - \frac{5}{3}}{-1} = \frac{y - 3}{2} = \frac{z - \frac{13}{3}}{-1}$

Step 8: Check the given options.

• Why: We substitute the coordinates of each option into the equation of the LSD to determine which point lies on it. A point lies on the line if, when substituted, all three ratios are equal to a common constant.

Let's test option (A): $\left(\frac{14}{3},-3, \frac{22}{3}\right)$ For $x$: $\frac{\frac{14}{3} - \frac{5}{3}}{-1} = \frac{\frac{9}{3}}{-1} = \frac{3}{-1} = -3$ For $y$: $\frac{-3 - 3}{2} = \frac{-6}{2} = -3$ For $z$: $\frac{\frac{22}{3} - \frac{13}{3}}{-1} = \frac{\frac{9}{3}}{-1} = \frac{3}{-1} = -3$ Since all three ratios are equal to $-3$, point (A) lies on the line of shortest distance.

Let's quickly verify other options:

• For (B) $\left(2,3, \frac{1}{3}\right)$: $\frac{2 - \frac{5}{3}}{-1} = -\frac{1}{3}$, but $\frac{3 - 3}{2} = 0$. Ratios are not equal.
• For (C) $\left(\frac{8}{3},-1, \frac{1}{3}\right)$: $\frac{\frac{8}{3} - \frac{5}{3}}{-1} = -1$, but $\frac{-1 - 3}{2} = -2$. Ratios are not equal.
• For (D) $\left(-\frac{5}{3},-7,1\right)$: $\frac{-\frac{5}{3} - \frac{5}{3}}{-1} = \frac{10}{3}$, but $\frac{-7 - 3}{2} = -5$. Ratios are not equal.

Thus, only option (A) lies on the line of shortest distance.

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Common Mistakes & Tips

• Cross Product Accuracy: Errors in calculating the cross product $\vec{b_1} \times \vec{b_2}$ are common. Double-check the determinant calculation, especially the signs, as it determines the direction of the LSD.
• Algebraic Precision: Solving the system of two linear equations for $\lambda$ and $\mu$ is a crucial step. Careless arithmetic mistakes at this stage will invalidate the entire solution.
• Understanding Perpendicularity: Remember that the shortest distance vector $\vec{PQ}$ must be perpendicular to both direction vectors $\vec{b_1}$ and $\vec{b_2}$. This yields two distinct dot product equations.

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Summary

To find a point on the line of shortest distance (LSD) between two skew lines, we first represent the lines in vector form. The direction vector of the LSD is obtained by taking the cross product of the direction vectors of the given lines, due to its perpendicularity to both. We then define general points on each line using parameters $\lambda$ and $\mu$, and form the vector connecting them ($\vec{PQ}$). By applying the perpendicularity conditions ($\vec{PQ} \cdot \vec{b_1} = 0$ and $\vec{PQ} \cdot \vec{b_2} = 0$), we solve for $\lambda$ and $\mu$. These values define the specific points (feet of the common perpendicular) on each line. Using one of these points and the LSD's direction vector, we formulate the Cartesian equation of the LSD. Finally, we substitute the coordinates of the given options into this equation to identify the point that lies on the line. Following this systematic approach, option (A) was found to be the correct answer.

The final answer is $\boxed{\text{(A)}}$.