Question

Two lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ are perpendicular to each other if.
$\begin{aligned} & a)aa'+cc'=-1 \\\ & b)aa'+cc'=1 \\\ & c)\dfrac{a}{a'}-\dfrac{c}{c'}=-1 \\\ & d)\dfrac{a}{a'}-\dfrac{c}{c'}=1 \\\ \end{aligned}$

Answer 1

Now first we will consider the two given lines and rearrange it in a way so that we get the equation of line in general Cartesian form which is $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ . Now we know that the lines $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ and $\dfrac{x-{{x}_{2}}}{a'}=\dfrac{y-{{y}_{2}}}{b'}=\dfrac{z-{{z}_{2}}}{c'}$ are perpendicular if aa’ + bb’ + cc’ = 0. Hence we will use the condition to find the required condition so that the given lines are perpendicular.

Complete step by step answer:
Now first let us consider the given line x = ay + b, z = cy + d
Now let us rearrange the terms in the above equation. Hence, we write it as $\dfrac{x-b}{a}=y$ and $\dfrac{z-d}{c}=y$ .
Hence we get the equation of the given line is $\dfrac{x-b}{a}=\dfrac{y}{1}=\dfrac{z-d}{c}..................\left( 1 \right)$
Now consider the given line x = a’y + b’, z = c’y + d’
Now let us again rearrange the terms in the above equation. Hence, we can write it as $\dfrac{x-b'}{a'}=y$ and $y=\dfrac{z-d'}{c'}$ .

Hence we can write the equation of line as $\dfrac{x-b'}{a'}=\dfrac{y}{1}=\dfrac{z-d'}{c'}...............\left( 2 \right)$
Now we know that the lines $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ and $\dfrac{x-{{x}_{2}}}{a'}=\dfrac{y-{{y}_{2}}}{b'}=\dfrac{z-{{z}_{2}}}{c'}$ are perpendicular if aa’ + bb’ + cc’ = 0
Hence now using this result we will find the condition such that the lines represented in equation (1) and equation (2) are perpendicular.
$aa'+1+cc'=0$
Now rearranging the terms of the above equation we get,
$aa'+cc'=-1$

So, the correct answer is “Option a”.

Note: Now we know that the equation of line in Cartesian form is $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$ .
We can convert the lines in vector form which will be $\vec{r}=\left( {{x}_{1}}\hat{i}+{{x}_{2}}\hat{j}+{{x}_{3}}\hat{k} \right)+\lambda \left( a\hat{i}+b\hat{j}+c\hat{k} \right)$ .
Now we know that the lines $\vec{r}=\vec{a}+{{\lambda }_{1}}\vec{b}$ and $\vec{r}=\vec{c}+{{\lambda }_{2}}\vec{d}$ are perpendicular if $\vec{b}.\vec{d}=0$ .
Hence we can also solve this by using the vector form of the equations.