Question

Points \( P \) and \( Q \) are given by \( \vec{OP} = \vec{i} - \vec{j} - \vec{k} \) and \( \vec{OQ} = -\vec{i} + \vec{j} + \vec{k} \). A line along the vector \( \vec{a} = \vec{i} + \vec{j} \) passes through point \( P \), and another line along the vector \( \vec{b} = \vec{j} - \vec{k} \) passes through point \( Q \). If a line along the vector \( \vec{c} = \vec{i} - \vec{j} + \vec{k} \) intersects both the lines along \( \vec{a} \) and \( \vec{b} \) at \( L \) and \( M \) respectively, then \( \vec{PM} = \) ?

• A. \( \vec{i} - \vec{j} + 2\vec{k} \)
• B. \( 4\vec{i} + 4\vec{j} \)
• C. \( -2\vec{i} + 10\vec{j} - 6\vec{k} \)
• D. \( 3\vec{i} - 2\vec{j} + \vec{k} \)

Hint:

To find the vector between two skew lines intersected by a third, express the points on each line parametrically and solve for the parameter values that match the third line's direction vector.

Answer 1

The Correct Option is C

Let us define the parametric equations of the three lines: Line through \( P \) and parallel to \( \vec{a} = \vec{i} + \vec{j} \): 
\[ \vec{r}_1(t) = (\vec{i} - \vec{j} - \vec{k}) + t(\vec{i} + \vec{j}) = (1 + t)\vec{i} + (-1 + t)\vec{j} - \vec{k} \] Line through \( Q \) and parallel to \( \vec{b} = \vec{j} - \vec{k} \): 
\[ \vec{r}_2(s) = (-\vec{i} + \vec{j} + \vec{k}) + s(\vec{j} - \vec{k}) = -\vec{i} + (1 + s)\vec{j} + (1 - s)\vec{k} \] Line along \( \vec{c} = \vec{i} - \vec{j} + \vec{k} \): 
Let this line intersect the above two lines at points \( L \) and \( M \), with parameter \( \lambda \), so: \[ \vec{r}_c(\lambda) = \vec{r}_L = \vec{r}_M = \vec{r}_0 + \lambda(\vec{i} - \vec{j} + \vec{k}) \] Now, we equate this vector with each line to find the common point of intersection: Let \( \vec{r}_c(\lambda) = \vec{r}_1(t) = \vec{r}_2(s) \) From the equality: \[ (1 + t)\vec{i} + (-1 + t)\vec{j} - \vec{k} = -\vec{i} + (1 + s)\vec{j} + (1 - s)\vec{k} \]