Question

Show that the lines given by,
$\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect.
Also find the co-ordinates of the point of intersection.

Answer 1

The given lines are
$\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}=r$ (say) .... (i)
and $\frac{x+10}{-1}+\frac{y+1}{-3}=\frac{z-1}{4}=R$ (say) .... (ii)
Any point on line (i) is
$(-1-10r,-3-r,4+r)$
Any point on line (ii)
$(-10-R,-1-3R,1+4R)$
At the point of intersection,
$\Rightarrow -1-10r=-10-R$
$\Rightarrow -3-r=-1-3R$
$\Rightarrow 4+r=1+4R$
$10r-R=9$ .... (iii)
$r-3R=-2$ .... (iv)
$r-4R=-3$ .... (v)
Solving (iv) and (v), we get
$r=1$ and $R=1$
These values of $r$ and $R$ satisfy (iii),
Hence, the given lines intersect.
For point of intersection, putting the value of $r$ or $R$ in (i), we get it as $(-11,-4,5)$.