Question

The lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}$ are
(a) coincident
(b) skew
(c) intersecting
(d) parallel

Answer 1

(a) coincident

The equation of the given lines are

$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}...\left(1\right)$


$$\begin{gathered} \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \\ =\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}...\left(2\right) \end{gathered}$$

Thus, the two lines are parallel to the vector $\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$ and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

$$\begin{gathered} \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)\times \overrightarrow{b}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)\times \left(\hat{i}+2\hat{j}+3\hat{k}\right) \\ =\overrightarrow{0}\left[\because \overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0}\right] \end{gathered}$$

Since the distance between the two parallel line is 0, the given lines are coincident.