Question

The lines $x=ay+b,z=cy+d,x=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }$ are perpendicular if $a{a}^{\prime }+c{c}^{\prime }$ equal to

• A. $-1$
• B. $0$
• C. $1$
• D. none of these

Answer 1

The correct option is A $-1$

The equation of the lines are

$x=ay+b,z=cy+d$

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

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

$\therefore$ direction ratios of two lines are $a,1,c$ and $a^{\prime },1,c^{\prime }$

Thus the two lines are perpendicular if

$\left(a\right)\left(a^{\prime }\right)+\left(1\right)\left(1\right)+\left(c\right)\left(c^{\prime }\right)=0$

$\therefore a{a}^{\prime }+c{c}^{\prime }=-1$.