Question

The lines $x=ay+b$ and $z=cy+d$ and $x=a{}^{\prime }y+b{}^{\prime }$ and $z=c{}^{\prime }y+d$ will be perpendicular, if and only if

• A. $aa{}^{\prime }+bb{}^{\prime }+cc{}^{\prime }=0$
• B. $(a+a{}^{\prime })(b+b{}^{\prime })(c+c{}^{\prime })=0$
• C. $aa{}^{\prime }+cc{}^{\prime }+1=0$
• D. $aa{}^{\prime }+bb{}^{\prime }+cc{}^{\prime }=-1$

Answer 1

The correct option is C $aa{}^{\prime }+cc{}^{\prime }+1=0$

The equation of the lines are $x=ay+b,z=cy+d,x=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }$

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

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

the two lines are perpendicular if $\left(a\right)\left(a^{\prime }\right)+\left(1\right)\left(1\right)+\left(c\right)\left(c^{\prime }\right)=0$

$\Rightarrow$ $[a_1{a}_2+b_1{b}_2+c_1{c}_2=0]$

if $aa\text{'}+cc\text{'}+1=0$.