Question

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

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

Answer 1

The correct option is A $a{a}^{\prime }+c{c}^{\prime }+1=0$

The equation of the line are :

$\begin{array}{c}x=ay+b,z=cy+d\\ andx=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }\\ \Rightarrow \frac{x-b^{\prime }}{a^{\prime }}=\frac{y}{1}=\frac{z-d^{\prime }}{c^{\prime }}\end{array}$

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

The two lines are perpendicular if

$\begin{array}{c}{\left(a\right)}^{\prime }a+\left(1\right)\left(1\right)+\left(c\right)\left(c^{\prime }\right)=0\left[\because a_1{a}_2+b_1{b}_2+c_1{c}_2=0\right]\\ i.eifa{a}^{\prime }+c{c}^{\prime }+1=0\end{array}$