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

•

Step 1 : Understanding the Question:
We are given two skew lines in three-dimensional space, \(L_1\) and \(L_2\).
We are given the point \(P_1(2, 3, 4)\) on the line \(L_1\) which is closest to \(L_2\).
We need to find the point \(P_2\) on the line \(L_2\) which is closest to \(L_1\).

•

Step 2 : Key Formula or Approach:
The line segment joining the closest points on two skew lines is perpendicular to both lines.
Thus, the vector \(\vec{P_1P_2}\) connecting the closest point \(P_1\) on \(L_1\) to the closest point \(P_2\) on \(L_2\) must be perpendicular to the direction vector of \(L_2\).
The direction vector of \(L_2\) is obtained from the coefficients of the parameter in the equation of \(L_2\).
If two vectors \(\vec{u}\) and \(\vec{v}\) are perpendicular, their dot product is zero: \(\vec{u} \cdot \vec{v} = 0\).

•

Step 3 : Detailed Explanation:
Let us write down the parametric equations of the lines:
For \(L_1\):
\[ x = 2 + \lambda, \quad y = 3 + 2\lambda, \quad z = 4 + 3\lambda \] The direction vector of \(L_1\) is \(\vec{d}_1 = (1, 2, 3)\).
The point \(P_1(2, 3, 4)\) corresponds to \(\lambda = 0\) on \(L_1\).
For \(L_2\):
\[ x = 4 + \mu, \quad y = 4, \quad z = 4 + \mu \] where we use \(\mu\) as the parameter to avoid confusion with \(\lambda\).
The direction vector of \(L_2\) is \(\vec{d}_2 = (1, 0, 1)\).
Any arbitrary point \(P_2\) on \(L_2\) can be written as:
\[ P_2 = (4 + \mu, 4, 4 + \mu) \] Let us find the vector connecting \(P_1(2, 3, 4)\) and \(P_2\):
\[ \vec{P_1P_2} = P_2 - P_1 = (4 + \mu - 2, 4 - 3, 4 + \mu - 4) = (2 + \mu, 1, \mu) \] Since \(P_1\) and \(P_2\) are the closest points, the vector \(\vec{P_1P_2}\) must be perpendicular to the direction vector of \(L_2\).
Therefore:
\[ \vec{P_1P_2} \cdot \vec{d}_2 = 0 \] \[ (2 + \mu)(1) + (1)(0) + (\mu)(1) = 0 \] \[ 2 + \mu + \mu = 0 \implies 2\mu = -2 \implies \mu = -1 \] Now, substitute \(\mu = -1\) into the coordinates of \(P_2\):
\[ P_2 = (4 - 1, 4, 4 - 1) = (3, 4, 3) \] Let us also verify if \(\vec{P_1P_2}\) is perpendicular to the direction vector of \(L_1\):
For \(\mu = -1\)., the vector is \(\vec{P_1P_2} = (1, 1, -1)\).
The dot product with \(\vec{d}_1 = (1, 2, 3)\) is:
\[ \vec{P_1P_2} \cdot \vec{d}_1 = 1(1) + 1(2) - 1(3) = 1 + 2 - 3 = 0 \] Since the vector is perpendicular to both lines, \((3, 4, 3)\) is indeed the unique closest point.

•

Step 4 : Final Answer:
The point on \(L_2\) closest to \(L_1\) is \((3, 4, 3)\).
This corresponds to option (A).