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

Answer 1

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

Line $x=ay+b,z=cy+d\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$

Line $x=a^{\prime }z+b^{\prime },y=c^{\prime }z+d^{\prime }$

$\to \frac{x-b^{\prime }}{a^{\prime }}=\frac{y-d^{\prime }}{c^{\prime }}=\frac{z}{1}$

Given both line are perpendicular

$\Rightarrow a{a}^{\prime }+c^{\prime }+c=0$