Question

Consider the lines $L_1$ and $L_2$ given below:
$L_1 : x = 2 + \lambda, \quad y = 3 + 2\lambda, \quad z = 4 + 3\lambda$;
$L_2 : x = 4 + \lambda, \quad y = 4, \quad z = 4 + \lambda$.
If $(2, 3, 4)$ is the point of $L_1$ that is closest to $L_2$, then which point of $L_2$ is closest to $L_1$?

• A. (3, 4, 3)
• B. (3, 4, 4)
• C. (5, 4, 5)
• D. (4, 4, 4)

Hint:

The shortest distance vector between two skew lines is always perpendicular to the direction vectors of both lines.
Finding the dot product of the parametric vector with the direction vector of the second line is a robust and straightforward way to find the unknown parameter.

Answer 1

The Correct Option is A

To determine the point on line \(L_2\) which is closest to line \(L_1\), we follow these steps:

Step 1: Parameterize the Lines

The equations for the lines are given as:

• \(L_1: x = 2 + \lambda, \quad y = 3 + 2\lambda, \quad z = 4 + 3\lambda\)
• \(L_2: x = 4 + \lambda, \quad y = 4, \quad z = 4 + \lambda\)

Both lines are parameterized with a parameter \(\lambda\).

Step 2: Find the Closest Point on \(L_2\) to \(L_1\)

Given that the point \((2, 3, 4)\) on \(L_1\) is closest to \(L_2\), we need to find the corresponding closest point on \(L_2\). For two lines \(L_1\) and \(L_2\), the closest points have perpendicular distance vectors.

Since the line \(L_2\) is given in terms of \(\lambda\) as

• \(x = 4 + \lambda\)
• \(y = 4\)
• \(z = 4 + \lambda\)

To find the parameter \(\lambda\) which gives the closest point on \(L_2\) to \((2, 3, 4)\), we set the directional vector of each line such that they are perpendicular.

The directional vector of \(L_1\) is \((1, 2, 3)\) and of \(L_2\) is \((1, 0, 1)\). The vector joining the point \((2, 3, 4)\) on \(L_1\) to any point on \(L_2\) given by \((4 + \lambda, 4, 4 + \lambda)\) is:

• \(((4 + \lambda) - 2, 4 - 3, (4 + \lambda) - 4) = (\lambda + 2, 1, \lambda)\)

This vector must be perpendicular to both \(L_1\) and \(L_2\), resulting in a zero dot product. Solving for this condition involves:

• \((\lambda + 2, 1, \lambda) \cdot (1, 0, 1) = (\lambda + 2) + 0 + \lambda = 0\)

Simplifying gives \(2\lambda + 2 = 0 \Rightarrow \lambda = -1\).

Substitute back into \(L_2\) gives:

• \(x = 4 - 1 = 3\)
• \(y = 4\)
• \(z = 4 - 1 = 3\)

Thus, the closest point on \(L_2\) is \((3, 4, 3)\).

Conclusion:

Therefore, the point on \(L_2\) that is closest to \(L_1\) is \((3, 4, 3)\).