Question

Two lines \[{L_1}:x = 5,\dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}\] and \[{L_2}:x = \alpha ,\dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}\] are coplanar. Then \[\alpha \] can take value(s)
This question has multiple correct options
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: In this question we will proceed by rewriting the given lines into their standard equation form. Then use the condition of coplanarity for two lines to be in coplanar to get the required values of \[\alpha \].

 Complete step by step answer:
Given lines are \[{L_1}:x = 5,\dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}\] and \[{L_2}:x = \alpha ,\dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}\]
Here we can rewrite lines \[{L_1}\] as \[{L_1}:\dfrac{{x - 5}}{0} = \dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}\] and the line \[{L_2}\] as \[{L_2}:\dfrac{{x - \alpha }}{0} = \dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}\] in the standard form.
We know that the condition of the two lines \[\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}\] and \[\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}\] to be coplanar is given by \[\left| {\begin{array}{*{20}{c}}
  {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\] or \[\left| {\begin{array}{*{20}{c}}
  {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\].
Since the given lines \[{L_1}\] and \[{L_2}\]are coplanar, according to the condition of coplanarity we have
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {x - 5}&y&z \\
  0&{3 - \alpha }&{ - 2} \\
  0&{ - 1}&{2 - \alpha }
\end{array}} \right| = 0\]
Opening the determinant with first column, we get
\[
   \Rightarrow x - 5\left[ {\left( {3 - \alpha } \right)\left( {2 - \alpha } \right) - \left( { - 2} \right)\left( { - 1} \right)} \right] - 0\left[ {y\left( {2 - \alpha } \right) - \left( { - 1} \right)\left( z \right)} \right] + 0\left[ {y\left( { - 2} \right) - z\left( { - 1} \right)} \right] = 0 \\
   \Rightarrow x - 5\left[ {\left( {{\alpha ^2} - 5\alpha + 6} \right) - 2} \right] - 0 + 0 = 0 \\
   \Rightarrow x - 5\left( {{\alpha ^2} - 5\alpha + 6 - 2} \right) = 0 \\
   \Rightarrow {\alpha ^2} - 5\alpha + 4 = \dfrac{0}{{x - 5}} \\
   \Rightarrow {\alpha ^2} - 5\alpha + 4 = 0 \\
\]
Splitting the terms and taking common, we have
\[
   \Rightarrow {\alpha ^2} - \alpha - 4\alpha + 4 = 0 \\
   \Rightarrow \alpha \left( {\alpha - 1} \right) - 4\left( {\alpha - 1} \right) = 0 \\
   \Rightarrow \left( {\alpha - 1} \right)\left( {\alpha - 4} \right) = 0 \\
  \therefore \alpha = 1,{\text{4}} \\
\]
Therefore, the values of \[\alpha \] are 1 and 4.
Thus, the correct options are A and D.

Note:
The condition that the two lines in three-dimensional \[\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}}\] and \[\dfrac{{x - {x_2}}}{{{q_1}}} = \dfrac{{y - {y_2}}}{{{q_2}}} = \dfrac{{z - {z_2}}}{{{q_3}}}\] to be coplanar is given by \[\left| {\begin{array}{*{20}{c}}
  {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\] or \[\left| {\begin{array}{*{20}{c}}
  {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\
  {{p_1}}&{{p_2}}&{{p_3}} \\
  {{q_1}}&{{q_2}}&{{q_3}}
\end{array}} \right| = 0\].