Question

The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
(a) 3, 1, −2

(b) 2, −4, 1

(c) $\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$

(d) $\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$

Answer 1

(a) 3, 1, −2

We have
x − y + z − 5 = 0 = x − 3y − 6

$$\begin{gathered} \Rightarrow x-3y-6=0 \\ x-y+z-5=0 \\ \Rightarrow x=3y+6...\left(1\right) \\ x-y+z-5=0...\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} 3y+6-y+z-5=0 \\ \Rightarrow 2y+z+1=0 \\ \Rightarrow y=\frac{-z-1}{2} \\ y=\frac{x-6}{3}\left[\mathrm{From}\left(1\right)\right] \\ \therefore \frac{x-6}{3}=y=\frac{-z-1}{2} \end{gathered}$$

So, the given equation can be re-written as

$\frac{x-6}{3}=\frac{y}{1}=\frac{z+1}{-2}$

Hence, the direction ratios of the given line are proportional to 3, 1,$-$2.