Question

The direction cosines of the line $\frac{3 x+1}{-3}=\frac{3 y+2}{6}=\frac{z}{-1}$ are

(a) $\left(\frac{1}{3}, \frac{2}{3}, 0\right)$

(b) $\left(-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$

(c) $\left(-\frac{1}{2}, 1,-\frac{1}{2}\right)$

(d) $\left(-\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$

Answer 1

Given equation can be rewritten as

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

$\therefore$ Direction cosines are $\left(-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right)$

So, The correct option of this question will be (B).