Question

If the lines $x=ay+b$ , $z=cy+d$ and $x=a'z+b'$ , $y=c'z+d'$ are perpendicular lines then which of the following options is true
$\begin{align}
  & \text{a) cc }\!\!'\!\!\text{ +a+a }\!\!'\!\!\text{ =0} \\
 & \text{b) aa }\!\!'\!\!\text{ +c+c }\!\!'\!\!\text{ =0} \\
 & \text{c) ab }\!\!'\!\!\text{ +bc }\!\!'\!\!\text{ +1=0} \\
 & \text{d) bb }\!\!'\!\!\text{ +cc }\!\!'\!\!\text{ +1=0} \\
\end{align}$

Answer 1

Hint: We know that if lines $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ and the line $\dfrac{x-{{x}_{2}}}{p} = \dfrac{y-{{y}_{2}}}{q} = \dfrac{z-{{z}_{2}}}{r}$ are perpendicular then we know that $ap+bq+cr=0$. Hence we will write the given equation in this form and then find the condition for lines to be perpendicular

Complete step-by-step solution:
Now consider the line $x=ay+b$ , $z=cy+d$ we want to write this line in the form of $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ . Hence we will rearrange the terms of the equation to do so.
Let $x=ay+b...........(1)$
And $z=cy+d...............(2)$
Now consider equation (1)
$x=ay+b$
Taking b on LHS we get
$x-b=ay$.
Now dividing the whole equation by ‘a’ we get.
$\dfrac{x-b}{a}=y........................(3)$
Now consider equation (2)
$z=cy+d$
Taking d on RHS we get
$z-d=cy$
Now dividing the whole equation by x we get
$\dfrac{z-d}{c}=y........................(4)$
Now from equation (3) and equation (4) we get
$\dfrac{x-b}{a}=\dfrac{z-d}{c}=\dfrac{y}{1}.....................(5)$
Hence now we know that the equation is in the form $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$
Now consider the line$x=a'z+b'$ , $y=c'z+d'$
Let $x=a'z+b'.....................(6)$
And $y=cz'+d'.................(7)$
Now consider equation (6)
$x=a'z+b'$
Now taking b’ to LHS we get.
$x-b'=a'z$
Now we will divide the equation by z.
We get $\dfrac{x-b'}{a'}=z.................(8)$
Now consider the equation (7)
$y=c'z+d'$
Taking d’ to LHS we get
$y-d'=c'z$
Now dividing the whole equation by c’ we get
$\dfrac{y-d'}{c'}=z.................(9)$
Now from equation (8) and equation (9) we get
$\dfrac{x-b'}{a'}=\dfrac{y-d'}{c'}=\dfrac{z}{1}..................(10)$
Hence we have the equation in the form $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$
Now we know that if the line $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ and the line $\dfrac{x-{{x}_{2}}}{p}+\dfrac{y-{{y}_{2}}}{q}+\dfrac{z-{{z}_{2}}}{r}$ are perpendicular then we know that $ap+bq+cr=0$.
Hence using this property to equation(5) and equation (10) we get
$aa'+c+c'=0$
Option b is the correct option.

Note: Now consider two lines $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ and the line $\dfrac{x-{{x}_{2}}}{p}=\dfrac{y-{{y}_{2}}}{q} =\dfrac{z-{{z}_{2}}}{r}$
Then a, b, c and p, q, r are the direction ratios of the lines respectively. Hence if the lines are perpendicular the dot product of the direction ratios of the lines is equal to zero. And hence we get the condition $ap+bq+cr=0$ .