Question

Let \( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5} \) be two lines. Then which of the following points lies on the line of the shortest distance between \( L_1 \) and \( L_2 \)?

• A. \( \left( \frac{-5}{3}, -7, 1 \right) \)
• B. \( (2, 3, \frac{1}{3}) \)
• C. \( \left( \frac{8}{3}, -1, \frac{1}{3} \right) \)
• D. \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \)

Hint:

The shortest distance between two skew lines is given by the line joining the points of intersection on each line.

Answer 1

The Correct Option is D

Solution to the Shortest Distance Between Two Lines 

We are given that line \( PQ \) is the line of shortest distance between the lines \( L_1 \) and \( L_2 \), where:

• Point \( P \) lies on \( L_1 \) and is of the form \( P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \).
• Point \( Q \) lies on \( L_2 \) and is of the form \( Q(3\mu + 2, 4\mu + 4, 5\mu + 5) \).

Step 1: Direction Ratios of \( PQ \)

The direction ratios of \( PQ \) are given as: \[ \text{Direction Ratios of } PQ = (3\mu - 2\lambda + 1, 4\mu - 3\lambda + 2, 5\mu - 4\lambda + 2) \] The condition \( PQ \perp L_2 \) (i.e., \( PQ \) is perpendicular to \( L_2 \)) implies that the dot product of the direction ratios of \( PQ \) and \( L_2 \) is zero.

Step 2: Set up the first equation using the dot product

The dot product of the direction ratios of \( PQ \) with those of \( L_2 \) gives the following equation: \[ (3\mu - 2\lambda + 1) \times 2 + (4\mu - 3\lambda + 2) \times 3 + (5\mu - 4\lambda + 2) \times 4 = 0 \] Simplifying the equation: \[ 2(3\mu - 2\lambda + 1) + 3(4\mu - 3\lambda + 2) + 4(5\mu - 4\lambda + 2) = 0 \] \[ 6\mu - 4\lambda + 2 + 12\mu - 9\lambda + 6 + 20\mu - 16\lambda + 8 = 0 \] \[ 38\mu - 29\lambda + 16 = 0 \quad \text{...(1)} \]

Step 3: Set up the second equation

Similarly, using the condition \( PQ \perp L_2 \), we get the second equation: \[ (3\mu - 2\lambda + 1) \times 3 + (4\mu - 3\lambda + 2) \times 4 + (5\mu - 4\lambda + 2) \times 5 = 0 \] Simplifying: \[ 3(3\mu - 2\lambda + 1) + 4(4\mu - 3\lambda + 2) + 5(5\mu - 4\lambda + 2) = 0 \] \[ 9\mu - 6\lambda + 3 + 16\mu - 12\lambda + 8 + 25\mu - 20\lambda + 10 = 0 \] \[ 50\mu - 38\lambda + 21 = 0 \quad \text{...(2)} \]

Step 4: Solve the system of equations

Now, we solve the system of equations: \[ 38\mu - 29\lambda + 16 = 0 \quad \text{...(1)} \] \[ 50\mu - 38\lambda + 21 = 0 \quad \text{...(2)} \] Using equation (1) and (2), solve for \( \lambda \) and \( \mu \): \[ \lambda = \frac{1}{3}, \quad \mu = \frac{-1}{6} \]

Step 5: Find the coordinates of points \( P \) and \( Q \)

Substituting \( \lambda = \frac{1}{3} \) and \( \mu = \frac{-1}{6} \) into the equations for \( P \) and \( Q \): \[ P\left( 2\left(\frac{1}{3}\right) + 1, 3\left(\frac{1}{3}\right) + 2, 4\left(\frac{1}{3}\right) + 3 \right) = \left( \frac{5}{3}, 3, \frac{13}{3} \right) \] \[ Q\left( 3\left(\frac{-1}{6}\right) + 2, 4\left(\frac{-1}{6}\right) + 4, 5\left(\frac{-1}{6}\right) + 5 \right) = \left( \frac{3}{2}, \frac{10}{3}, \frac{25}{6} \right) \]

Step 6: Equation of the line \( PQ \)

The parametric form of the line \( PQ \) is given by: \[ \frac{x - \frac{5}{3}}{\frac{1}{6}} = \frac{y - 3}{\frac{-1}{3}} = \frac{z - \frac{13}{3}}{\frac{1}{6}} \] Simplifying this, we get: \[ \frac{x - \frac{5}{3}}{1} = \frac{y - 3}{-2} = \frac{z - \frac{13}{3}}{1} \]

Step 7: Verify that the point \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) lies on the line \( PQ \)

Substituting the coordinates \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) into the parametric equations: \[ \frac{x - \frac{5}{3}}{1} = \frac{14/3 - 5/3}{1} = \frac{9}{3} = 3 \] \[ \frac{y - 3}{-2} = \frac{-3 - 3}{-2} = \frac{-6}{-2} = 3 \] \[ \frac{z - \frac{13}{3}}{1} = \frac{22/3 - 13/3}{1} = \frac{9}{3} = 3 \] Since all equations are equal, the point \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) lies on the line \( PQ \).

Conclusion

Therefore, the point \( \left( \frac{14}{3}, -3, \frac{22}{3} \right) \) lies on the line \( PQ \), confirming that our calculations are correct.

Answer 2

Finding the Point on the Line of Shortest Distance Between Two Lines

We are given two lines:

\( L_1: \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4} \)

\( L_2: \dfrac{x - 2}{3} = \dfrac{y - 4}{4} = \dfrac{z - 5}{5} \)

Step 1: Parametric Equations of Lines

Let the parameter for \( L_1 \) be \( \lambda \) and for \( L_2 \) be \( \mu \).

Then,

\[ L_1: \begin{cases} x = 2\lambda + 1 \\ y = 3\lambda + 2 \\ z = 4\lambda + 3 \end{cases} \quad L_2: \begin{cases} x = 3\mu + 2 \\ y = 4\mu + 4 \\ z = 5\mu + 5 \end{cases} \]

Step 2: Coordinates of Points on \( L_1 \) and \( L_2 \)

Let \( P \) be a point on \( L_1 \) and \( Q \) be a point on \( L_2 \):

\[ P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \] \[ Q(3\mu + 2, 4\mu + 4, 5\mu + 5) \]

Step 3: Direction Ratios of Line \( PQ \)

The direction ratios of \( PQ \) are given by:

\[ PQ = (3\mu - 2\lambda + 1,\ 4\mu - 3\lambda + 2,\ 5\mu - 4\lambda + 2) \]

Since \( PQ \) is perpendicular to both \( L_1 \) and \( L_2 \), we use the dot product condition:

\[ PQ \cdot (2,3,4) = 0 \quad \text{and} \quad PQ \cdot (3,4,5) = 0 \]

Step 4: Forming the Two Equations

From the first condition:

\[ 2(3\mu - 2\lambda + 1) + 3(4\mu - 3\lambda + 2) + 4(5\mu - 4\lambda + 2) = 0 \] \[ 6\mu - 4\lambda + 2 + 12\mu - 9\lambda + 6 + 20\mu - 16\lambda + 8 = 0 \] \[ 38\mu - 29\lambda + 16 = 0 \quad \text{...(1)} \]

From the second condition:

\[ 3(3\mu - 2\lambda + 1) + 4(4\mu - 3\lambda + 2) + 5(5\mu - 4\lambda + 2) = 0 \] \[ 9\mu - 6\lambda + 3 + 16\mu - 12\lambda + 8 + 25\mu - 20\lambda + 10 = 0 \] \[ 50\mu - 38\lambda + 21 = 0 \quad \text{...(2)} \]

Step 5: Solving the Two Equations

\[ 38\mu - 29\lambda + 16 = 0 \] \[ 50\mu - 38\lambda + 21 = 0 \]

Solving these simultaneously gives:

\[ \lambda = \frac{1}{3}, \quad \mu = -\frac{1}{6} \]

Step 6: Coordinates of \( P \) and \( Q \)

For \( \lambda = \frac{1}{3} \):

\[ P\left(2\left(\frac{1}{3}\right)+1,\ 3\left(\frac{1}{3}\right)+2,\ 4\left(\frac{1}{3}\right)+3\right) = \left(\frac{5}{3}, 3, \frac{13}{3}\right) \]

For \( \mu = -\frac{1}{6} \):

\[ Q\left(3\left(-\frac{1}{6}\right)+2,\ 4\left(-\frac{1}{6}\right)+4,\ 5\left(-\frac{1}{6}\right)+5\right) = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \]

Step 7: Equation of Line \( PQ \)

Direction ratios of \( PQ = Q - P = \left(\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right) \)

Hence, equation of \( PQ \) is:

\[ \frac{x - \frac{5}{3}}{1} = \frac{y - 3}{-2} = \frac{z - \frac{13}{3}}{1} \]

Step 8: Check the Given Points

Substitute \( \left(\frac{14}{3}, -3, \frac{22}{3}\right) \):

\[ \frac{x - \frac{5}{3}}{1} = \frac{14/3 - 5/3}{1} = 3 \] \[ \frac{y - 3}{-2} = \frac{-3 - 3}{-2} = 3 \] \[ \frac{z - \frac{13}{3}}{1} = \frac{22/3 - 13/3}{1} = 3 \]

Since all ratios are equal, the point lies on the line \( PQ \).

Final Answer:

The point \( \left(\frac{14}{3}, -3, \frac{22}{3}\right) \) lies on the line of shortest distance between \( L_1 \) and \( L_2 \).

Hence, the correct option is (4).