Question

The d.r.s of the lines $x=ay+b,\ z=cy+d$ are
(A) 1, a, c (B) a, 1, c (C) b, 1, c (D) c, a, 1

Answer 1

Try to remember that drs stands for direction ratios write the equation of the 3-D line in standard form of:
$\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n}$

Complete step-by-step answer:
Once you write these, l, m, and n will give you the direction ratios.
D.r.s stands for direction ratios, can also be termed direction numbers, in the 3D equation of a line, the direction ration are generally given by the denominator coefficients of the x, y and z terms
In this question, we have been given the following.
$x=ay+b$ $z=cy+d$
This can be rewritten as:
$\dfrac{x-b}{a}=y$ $\dfrac{z-d}{c}=y$
Hence the equation of the line in 3D becomes
$\dfrac{x-b}{a}={{y}_{1}}=\dfrac{z-d}{c}$
As mentioned before, the denominator terms indicate the direction ratios.
Hence, D.r.s are a, 1, c option B is correct.

Note: If a, b, c are 3 numbers proportional to the direction cosines l, m, n of a straight line, then they are called the line’s direction ratio.
So we have
$\dfrac{\text{l}}{\text{a}}\ \text{=}\ \dfrac{\text{m}}{\text{b}}\ \text{=}\ \dfrac{\text{n}}{\text{c}}\ \text{=}\ \text{k}\ \ \text{(say)}$
So $\text{l}=\text{ak,}\ \ \text{m}=\text{bk,}\ \ \text{n}=\text{ck}$
But we know ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$
So ${{k}^{2}}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)=1$ [sum of direction cosine squares $=1$]
$k=\pm \dfrac{1}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
And therefore
$l=\pm \ ak=\dfrac{\pm \ a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
$m=\pm \ bk=\dfrac{\pm \ b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
$n=\pm \ ck=\dfrac{\pm \ b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
The direction cosines of a line are unique but the direction ratio may be infinite in number.