Question

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

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

Answer 1

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

The two given lines are:-

$L_1\equiv x=ay+b,z=cy+d$

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

we can write $L_1$ as

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

Similarly,

$L_2\equiv x=a^{\prime }y+b^{\prime },z=c^{\prime }y+d^{\prime }$

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

$L_2$ can be written as:-

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

Direction ratios of line $L_1(a,c,1)$

and

Direction ratios of line $L_2(a^{\prime },c^{\prime },1)$

We know that two lines are perpendicular iff, product of their direction ratios are 0.

Therefore,

$a\times a^{\prime }+c\times c^{\prime }+1\times 1=0$

$\Rightarrow a{a}^{\prime }+c{c}^{\prime }+1=0$

Hence, C is the correct answer.