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Question: Two lines \[{L_1}:x = 5,\dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}\] and \[{L_2}:x = \alpha ,\dfra...

Two lines L1:x=5,y3−α=z−2{L_1}:x = 5,\dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}} and L2:x=α,y−1=z2−α{L_2}:x = \alpha ,\dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }} are coplanar. Then α\alpha can take value(s)
This question has multiple correct options
A. 1
B. 2
C. 3
D. 4

Explanation

Solution

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 α\alpha .

Complete step by step answer:
Given lines are L1:x=5,y3−α=z−2{L_1}:x = 5,\dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}} and L2:x=α,y−1=z2−α{L_2}:x = \alpha ,\dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}
Here we can rewrite lines L1{L_1} as L1:x−50=y3−α=z−2{L_1}:\dfrac{{x - 5}}{0} = \dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}} and the line L2{L_2} as L2:x−α0=y−1=z2−α{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 x−x1p1=y−y1p2=z−z1p3\dfrac{{x - {x_1}}}{{{p_1}}} = \dfrac{{y - {y_1}}}{{{p_2}}} = \dfrac{{z - {z_1}}}{{{p_3}}} and x−x2q1=y−y2q2=z−z2q3\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 L1{L_1} and L2{L_2}are coplanar, according to the condition of coplanarity we have

{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 \\

Splittingthetermsandtakingcommon,wehaveSplitting 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 $$\left| {\begin{array}{*{20}{c}} {x - {x_2}}&{y - {y_2}}&{z - {z_2}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right| = 0$$.