Question

If the lines $x=ay+b,z=cy+d$ and $x=a^{\prime }z+b^{\prime },y=c^{\prime }z+d^{\prime }$ are perpendicular, then :

• A. $a{b}^{\prime }+b{c}^{\prime }+1=0$
• B. $b{b}^{\prime }+c{c}^{\prime }+1=0$
• C. $a{a}^{\prime }+c+c^{\prime }=0$
• D. $c{c}^{\prime }+a+a^{\prime }=0$

Answer 1

The correct option is C $a{a}^{\prime }+c+c^{\prime }=0$

Consider the lines
$x=ay+b,z=cy+d$
$\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$
$\therefore \overrightarrow{b_1}=a\hat{i}+\hat{j}+c\hat{k}$

Now,
$x=a^{\prime }z+b^{\prime },y=c^{\prime }z+d^{\prime }$
$\Rightarrow \frac{x-b^{\prime }}{a^{\prime }}=\frac{y-d^{\prime }}{c^{\prime }}=\frac{z}{1}$
$\therefore \overrightarrow{b_2}=a^{\prime }\hat{i}+c^{\prime }\hat{j}+\hat{k}$

The two lines are perpendicular.
So, $\overrightarrow{b_1}\cdot \overrightarrow{b_2}=0$
$\Rightarrow a{a}^{\prime }+c^{\prime }+c=0$