Question

If the straight lines $\frac{x+1}{2}=\frac{-y+1}{3}=\frac{z+1}{-2}$ and $\frac{x-3}{1}=\frac{y-\lambda }{2}=\frac{z}{3}$ intersect, then the value of $\lambda$ is

• A. $-\frac{5}{8}$
• B. $-\frac{17}{8}$
• C. $-\frac{13}{8}$
• D. $-\frac{15}{8}$
• E. $-\frac{21}{8}$

Answer 1

The correct option is A $-\frac{5}{8}$

Given lines are,
$\frac{x+1}{2}=\frac{-y+1}{3}=\frac{z+1}{-2}=\left(r_1\right)\left(\text{say}\right)$ .... (i)
So, $(2r_1-1,1-3r_1,-2r_1-1)$ be any point on line (i)
and $\frac{x-3}{1}=\frac{y-\lambda }{2}=\frac{z}{3}=r_2\left(\text{say}\right)$ .... (ii)
So, $(r_2+3,2r_2+\lambda .3r_2)$ be any point on line (ii).
Since, both lines intersect each other.
Thus $2r_1-1=r_2+3$
$\Rightarrow 2r_1-r_2=4$ .... (iii)
$-3r_1-1=2r_2+\lambda$
$\Rightarrow -3f_1-2r_2=\lambda -1$ ..... (iv)
and $-2r_1-1=3r_2$
$\Rightarrow 2r_1+3r_2=-1$ ....(v)
On subtracting Eq. (v) from Eq. (iii), we get
$-4r_2=5\Rightarrow r_2=-5/4$
From Eq. (iii), we get
$2r_1=4-\frac{5}{4}=\frac{11}{4}$
$\Rightarrow r_1=\frac{11}{8}$
On putting the value of $r_1$ and $r_2$ in Eq. (iv), we get
$-3\left(\frac{11}{8}\right)-2\left(\frac{5}{4}\right)=\lambda -1$
$\Rightarrow \lambda -1=-\frac{33}{8}+\frac{20}{8}=\frac{-13}{8}$
Therefore, $\lambda =1-\frac{13}{8}=\frac{-5}{8}$