Discussion Forum

 If lines   $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$  and  $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$   intersect  each other, then $\lambda$=.....

Answer:

Explanation:

  Given lines are ,

 $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda_{1}$  (let)  ..........(i)

 and  $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}=\lambda_{2}$ (let) ......(ii)

 Then, any point on the lie (i) is

 $(2\lambda_{1}+1, 3\lambda_{1}-1,4\lambda_{1}+1)$   and any point on line (ii) is 

$(\lambda_{2}+3, 2\lambda_{2}+\lambda,\lambda_{2})$ 

 Clearly, then lines (i) and (ii) will intersect , if 

$(2\lambda_{1}+1, 3\lambda_{1}-1,4\lambda_{1}+1)$ = $(\lambda_{2}+3, 2\lambda_{2}+\lambda,\lambda_{2})$

 For some particular value of $\lambda_{1}$ and $\lambda_{2}$

 $\Rightarrow$   $2\lambda_{1}+1=\lambda_{2}+3, 3\lambda_{1}-1=2\lambda_{2}+\lambda, and 4\lambda_{1}+1=\lambda_{2}$

$\Rightarrow$   $2\lambda_{1}-\lambda_{2}=2,3\lambda_{1}-2\lambda_{2}=\lambda+1$   and$4\lambda_{1}-\lambda_{2}=-1$

 On solving $2\lambda_{1}-\lambda_{2}=2,4\lambda_{1}-\lambda_{2}=-1$

 We get,   $\lambda_{1}=-\frac{3}{2}$ and $\lambda_{2}=-5$

Now putting the values of $\lambda_{1}$ and $\lambda_{2}$ in 

 $3\lambda_{1}-2\lambda_{2}=\lambda+1$

 $\Rightarrow$   $3(\frac{-3}{2})-2(-5)=\lambda+1\Rightarrow \frac{-9}{2}+10=\lambda+1$

 $\Rightarrow$   $\lambda+1=\frac{11}{2}\Rightarrow\lambda=\frac{9}{2}$