Question

If the lines $\frac{x-1}{2}=\frac{y+3}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then the value of k is

• A. $\frac{3}{2}$
• B. $\frac{9}{2}$
• C. $-\frac{2}{9}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{9}{2}$

Answer 2

Since, the lines intersect, therefore they must have a point in common, i.e.
\hspace20mm \frac{x-1}{2} = \frac{y+1}{3}=\frac{z-1}{4} = \lambda
and \hspace15mm \frac{x-3}{1} = \frac{y-k}{2}=\frac{z}{1} = \mu
\Rightarrow\hspace15mm x=2 \lambda + 1,y = 3 \lambda -1
\hspace20mm z = 4 \lambda + 1
and $\, \, \, x = \mu +3, y = 2\, \mu + k,z = \mu$ are same.
\Rightarrow\hspace15mm 2 \lambda + 1 = \mu +3
\hspace20mm 3 \lambda - 1 = 2\mu + k
\hspace20mm 4 \lambda + 1 = \mu
On solving Ist and IIIrd terms, we get, $\lambda = - \frac{3}{2}$
and \hspace15mm \mu = -5
\therefore \hspace15mm k =3 \lambda - 2\mu - 1
\Rightarrow \hspace15mm k =3 \bigg(-\frac{3}{2}\bigg)-2(-5)-1 =\frac{9}{2}
\therefore \hspace15mm k =\frac{9}{2}