Find the direction ratio and direction cosines of a line parallel to…

The correct answer is Option (1) → DRs: $(1, 2, 3)$; DCs: $(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}})$ ##

Given line \(L\): 

$6x - 2 = 3y + 1 = 2z – 2$

or $\frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y - \left(-\frac{1}{3}\right)}{\frac{1}{3}} = \frac{z - 1}{\frac{1}{2}}$

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

So, d.r.'s of line \(L\) are \(1, 2, 3\).

∴ Direction cosines of line \(L\) are:

$\frac{1}{\sqrt{1^2 + 2^2 + 3^2}}, \frac{2}{\sqrt{1^2 + 2^2 + 3^2}}, \frac{3}{\sqrt{1^2 + 2^2 + 3^2}}$

i.e., $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

Now, since the parallel lines have the proportional d.r.'s and direction cosines, so d.r.'s of required line passing through \(\left( \frac{1}{3}, -\frac{1}{3}, 1 \right)\) are \(1, 2, 3\) and direction cosines are

$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$