Discussion Forum

 Two lines L₁ :x=5 , $\frac{y}{3-\alpha}=\frac{z}{-2}$    and   $L_{2}:x=\alpha,\frac{y}{-1}=\frac{z}{2-\alpha}$ are coplannar , Then , $\alpha$ can take value(s)

Answer:

Explanation:

Concept involved

 If two straight lines are coplannar

i.e,   $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$  and 

$\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ are coplannar

  Then ,(x₂-x₁ , y₂-y₁,z₂-z₁), (a₁,b₁,c₁)  and (a₂,b₂,c₂) are coplanar

 i,e,

   $\begin{bmatrix}x_{2}-x_{1} & y_{2}-y_{1}&z_{2}-z_{1} \\a_{1} & b_{1}&c_{1}\\a_{2}&b_{2}&c_{2} \end{bmatrix}=0$

Here  x=5, $\frac{y}{3-\alpha}=\frac{z}{-2}$

 $\Rightarrow \frac{x-5}{0}=\frac{y-0}{-(\alpha-3)}=\frac{z-0}{-2}$......(i)

 and   $x=\alpha,\frac{y}{-1}=\frac{z}{2-\alpha}$

  $\Rightarrow \frac{x-\alpha}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}$   .....(ii)

$\Rightarrow    \begin{bmatrix}5-\alpha & 0&0\\0 & 3-\alpha&-2\\0&-1&2-\alpha \end{bmatrix}=0$

  $\Rightarrow$     $(5-\alpha)[(3-\alpha)(2-\alpha)-2]=0$

 $\Rightarrow$    $(5-\alpha)[\alpha^{2}-5\alpha+4]=0$

 $\Rightarrow$    $(5-\alpha)(\alpha-1)(\alpha-4)=0$

                     $\alpha$  =1,4,5