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Q.
If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then the value of $k$ is
Three Dimensional Geometry
Solution:
Since, the lines intersect they must have a point in common i.e.,
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$
$\Rightarrow x=2\lambda+1$,
$y=3\lambda-1$,
$z=4\lambda+1$
and $x=\mu+3$,
$y=2\mu+k$,
$z=\mu$
$\Rightarrow 2\lambda+1=\mu+3, 3\lambda-1=2\mu+k, 4\lambda+1=\mu$
On solving we get,
$\lambda=-\frac{3}{2}$ and $\mu=-5$
$\therefore k=3\lambda-2\mu-1=3\left(-\frac{3}{2}\right)-2\left(-5\right)-1$
$\Rightarrow k=\frac{9}{2}$