Question

The two lines $x-ay+b,z=cy+d$ and $x=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }$ are perpendicular to each other, if:

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

Answer 1

Answer: (D)

The equations of given lines are
$x=ay+bjZ=cy+d$
and $x=a^{\prime }y+b^{\prime },z=c’y+d’$
These equations can be rewritten as
$\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}$
and $\frac{x-b^{\prime }}{a^{\prime }}=\frac{y-0}{1}=\frac{z-d^{\prime }}{c^{\prime }}$
These lines will perpendicular, if
$a{a}^{\prime }+c{c}^{\prime }+1=0$