Question

If the lines $\dfrac{{{\text{x - 2}}}}{1} = \dfrac{{{\text{y - 3}}}}{1} = \dfrac{{{\text{z - 4}}}}{{ - {\text{k}}}}$and $\dfrac{{{\text{x - 1}}}}{{\text{k}}} = \dfrac{{{\text{y - 4}}}}{2} = \dfrac{{{\text{z - 5}}}}{1}$are coplanar, then k can have
$
  {\text{A}}{\text{. exactly one value}} \\
  {\text{B}}{\text{. exactly two values}} \\
  {\text{C}}{\text{. exactly three values}} \\
  {\text{D}}{\text{. any value}} \\
$

Answer 1

Hint: In order to find the number of values k can have, we consider the given line equations with three coordinates x, y and z. Given those lines are coplanar, i.e. they lie on the same plane. We find the directions of both the given lines and the direction of the line joining them. Then we compute their determinant of these three vectors to determine the value of k.

Complete step-by-step answer:
Given Data,
$\dfrac{{{\text{x - 2}}}}{1} = \dfrac{{{\text{y - 3}}}}{1} = \dfrac{{{\text{z - 4}}}}{{ - {\text{k}}}}$
$\dfrac{{{\text{x - 1}}}}{{\text{k}}} = \dfrac{{{\text{y - 4}}}}{2} = \dfrac{{{\text{z - 5}}}}{1}$

Let us consider the given lines as Line – 1 and Line – 2 respectively.
Let L – 1: $\dfrac{{{\text{x - 2}}}}{1} = \dfrac{{{\text{y - 3}}}}{1} = \dfrac{{{\text{z - 4}}}}{{ - {\text{k}}}}$
L – 2: $\dfrac{{{\text{x - 1}}}}{{\text{k}}} = \dfrac{{{\text{y - 4}}}}{2} = \dfrac{{{\text{z - 5}}}}{1}$

The direction of a line is given by its x, y and z co-ordinates respectively.
Therefore the direction of line – 1 is (1, 1, -k) and the direction of line – 2 is given by (k, 2, 1).

The positional vector of a point on a line is given by subtracting the respective value from the x, y and z co-ordinates.
Therefore the positional vector of a point on line – 1 is given by (2, 3, 4)
The positional vector of a point on line – 2 is given by (1, 4, 5)
The vector joining two lines of equations l - 1 and l – 2 is given by [(l – 1) – (l – 2)]
Therefore the vector joining our L – 1 and L – 2 is given by:
(2, 3, 4) – (1, 4, 5) = (1, -1, -1)

According to the question the lines L – 1 and L – 2 are co-planar, i.e. they lie on the same plane, so obviously the vector joining both of them should also be on the same plane.
We know if three vectors lie on the same plane, their determinant of their vector equations is equal to zero.
$
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&{ - 1}&{ - 1} \\
  1&1&{ - {\text{k}}} \\
  {\text{k}}&2&1
\end{array}} \right] = 0 \\
   \Rightarrow 1\left( {1 + 2{\text{k}}} \right) + 1\left( {1 + {{\text{k}}^2}} \right) - 1\left( {2 - {\text{k}}} \right) = 0 \\
   \Rightarrow {{\text{k}}^2} + 1 + 2{\text{k + 1 - 2 + k = 0}} \\
   \Rightarrow {{\text{k}}^2} + 3{\text{k = 0}} \\
   \Rightarrow {\text{k}}\left( {{\text{k + 3}}} \right) = 0 \\
   \Rightarrow {\text{k = 0 or k = - 3}} \\
$
Hence k has exactly two values.
Option B is the correct answer.

Note: In order to solve this type of problems the key is to know the concepts of lines in 3 – dimensional geometry, vector equations, positional vectors and their respective formula. We have to know the condition for co-planar lines and their definition. We have to know how to compute a determinant of a matrix to solve this type of problem.