Question

If for some $\mathrm{\alpha }\in \mathrm{R}$, the lines ${\mathrm{L}}_1:\frac{\left(\mathrm{x}+1\right)}{2}=\frac{\left(\mathrm{y}-2\right)}{-1}=\frac{(\mathrm{z}-1)}{1}$ and ${\mathrm{L}}_2:\frac{(\mathrm{x}+2)}{\mathrm{\alpha }}=\frac{\left(\mathrm{y}+1)\right)}{5-\mathrm{\alpha }}=\frac{(\mathrm{z}+1)}{1}$are coplanar, then the line ${\mathrm{L}}_2$ passes through the point

• A. $(2,-10,-2)$
• B. $(10,-2,-2)$
• C. $(10,2,2)$
• D. $(-2,10,2)$

Answer 1

The correct option is A

$(2,-10,-2)$

Step 1:Given Information

$$\begin{gathered} {\mathrm{L}}_1:\frac{(\mathrm{x}+1)}{2}=\frac{(\mathrm{y}-2)}{-1}=\frac{(\mathrm{z}-1)}{1} \\ {\mathrm{L}}_2:\frac{(\mathrm{x}+2)}{\mathrm{\alpha }}=\frac{(\mathrm{y}+1)}{(5-\mathrm{\alpha })}=\frac{(\mathrm{z}+1)}{1} \end{gathered}$$

Representation of above equation in form of points,

${\mathrm{L}}_1:\mathrm{A}(-1,2,1)$lies on the line L₁ and L₁ parallel to $\overrightarrow{\mathrm{p}}=2\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}$

${\mathrm{L}}_2:\mathrm{B}(-2,-1,-1)$ lies on the line L₂ and L₂ parallel to $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{\alpha i}}+\overrightarrow{(5-\mathrm{\alpha })\mathrm{j}}+\overrightarrow{\mathrm{k}}$

Step 2: Representation of given equation in form of matrix

since, L₁ and L₂ are co-planes,

$\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}$perpendicular to plane and $\overrightarrow{\mathrm{BA}}$ lies on the same plane

$\left(\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}\right).\overrightarrow{\mathrm{B}}=0$

$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}1-2& -2+1& -1-1\\ 2& -1& 1\\ \mathrm{\alpha }& 5-\mathrm{\alpha }& 1\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}-1& -3& -2\\ 2& -1& 1\\ \mathrm{\alpha }& 5-\mathrm{\alpha }& 1\end{array}\right|=0 \\ \Rightarrow -1(-1+\mathrm{\alpha }-5)+3(2-\mathrm{\alpha })-2(10-2\mathrm{\alpha }+\mathrm{\alpha })=0 \\ \Rightarrow 6-\mathrm{\alpha }+6-3\mathrm{\alpha }+2\mathrm{\alpha }-20=0 \\ \Rightarrow -8-2\mathrm{\alpha }=0 \\ \Rightarrow \mathrm{\alpha }=-4 \end{gathered}$$

Step 3: Applying $\mathrm{\alpha }$ value in L₂

$$\begin{gathered} {\mathrm{L}}_2:\frac{\left(\mathrm{x}+2\right)}{-4}=\frac{\left(\mathrm{y}+1\right)}{5-(-4)}=\frac{\left(\mathrm{z}+1\right)}{1} \\ {\mathrm{L}}_2:\frac{\left(\mathrm{x}+2\right)}{-4}=\frac{\left(\mathrm{y}+1\right)}{9}=\frac{\left(\mathrm{z}+1\right)}{1} \end{gathered}$$

Therefore, substituting values $\mathrm{x}=2,\mathrm{y}=-10,\mathrm{z}=-2$ will satisfy the above equation, which shows the option $(2,-10,-2)$ satisfies the above equation.

Hence, the correct answer is option (A).