Discussion Forum

If the lines   $\frac{x-2}{1} =\frac{y-3}{1}=\frac{z-4}{-k}$  and   $\frac{x-1}{k} =\frac{y-4}{2}=\frac{z-5}{1}$ are coplannar , then k can have 

Answer:

Explanation:

Condition for two lines  are coplannar

  $\begin{bmatrix}x_{1}-x_{2} & y_{1}-y_{2} &z_{1}-z_{2} \\l_{1}& m_{1}&n_{1} \\ l_{2}&m_{2}&n_{2} \end{bmatrix}=0$

 where , (x₁,y₁,z₁) and (x₂,y₂,z₂)  are the points lie ona line (i) and (ii) respectively and <l₁,m₁,n₁ >  and     

< l_2,,m₂,n₂ >  are the direction cosine of the line (i) and (ii)  respectively.

              $\therefore$    $\begin{vmatrix}2-1 & 3-4 &4-5 \\1& 1&-k \\ k&2&1 \end{vmatrix}=0$

$\begin{vmatrix}1 & -1 &-1 \\1& 1&-k \\ k&2&1 \end{vmatrix}=0$

  $\Rightarrow$       $1(1+2k)+(1+k^{2})-(2-k)=0$           

$\Rightarrow$     $k^{2}+2k+k=0$   

$\Rightarrow$      $k^{2}+3k=0\Rightarrow k=0,-3$

 If 0 appears in the denominator , then the correct way of representing the equation of straight line is

    $\frac{x-2}{1}=\frac{y-3}{1};z=4$