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# The two lines whose equations are given by $x=ay+b,\:\:z=cy+d\:\:\:and\:\:\:x=a'y+b',\:\:z=c'y+d'$ are $\perp$ if

$\begin{array}{1 1} (a)\:\:aa'+bb'+cc'=0\:\:\:\qquad\:\:(b)\:\:aa'+cc'+1=0\:\:\:\qquad\:\:(c)\:\:aa'+bb'+cc'=1\:\:\:\qquad\:\:(d)\:\:aa'+bb'+cc'=-1 \end{array}$

Given eqns. of the lines are $x-ay-b=0,\:\:\:cy-z+d=0$..........(i) and
$x-a'y-b'=0,\:\:\:c'y-z+d'=0$............(ii)
The equations of the lines are given in the form of two equations of the planes
$i.e.,$ The lines are intersection of the given two planes respectively.
$\Rightarrow$Each line is $\perp$ to the normals to both the corresponding planes by which the line is formed.
$\Rightarrow$ $d.r.$ of the line (i) is $(1,-a,0)\times(0,c,-1)$ and
$d.r.$ of the line (ii) is $(1,-a',0)\times (0,c',-1)$
$i.e.,\: d.r.$ of (i) is $\left |\begin{array}{ccc} \hat i& \hat j & \hat k\\1 & -a & 0\\0 & c &-1\end {array}\right|=(a,1,c)$
Similarly $d.r.$ of (ii) is $(a',1,c')$
If the lines (i) and (ii) are $\perp$ then $(a,1,c).(a',1,c')=0$
$\Rightarrow\:aa'+1+cc'=0$