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

Answer 1

The correct option is A $aa{}^{\prime }+cc{}^{\prime }=-1$

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

$L_1:\frac{x-b}{a}=y=\frac{z-d}{c}$

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

$L_2:\frac{x-b^{\prime }}{a^{\prime }}=y=\frac{z-d^{\prime }}{c^{\prime }}$

If the two lines are perpendicular angle between their direction ratios is $\frac{\pi }{2}$

$\cos \frac{\pi }{2}=0=a{a}^{\prime }+c{c}^{\prime }+1$

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