Mathematics | VITEEE 2014 | Discussion Page

Two line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point , if k is equal to

$\frac{2}{9}$

$\frac{1}{2}$

$\frac{9}{2}$

$\frac{1}{6}$

Option C

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}= r$ (say)

$\Rightarrow$ $x=2r+1.y=3r-1,z=4r+1$

Since, the two lines intersect

So, putting above values in second lines, we get

$\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}$

$2r-2=4r+1$

$\Rightarrow$ $r=-3/2$

Also, $3r-1-k=8r+2$

$\Rightarrow$ $k=-5r-3=\frac{15}{2}-3=\frac{9}{2}$

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