The direction cosines of the line given by the equations \(x = 2y+3\) and \(z = 2y-1\) are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\)
Let \(y = t\). Then every point of the line is
\(x = 2t+3,\quad y = t,\quad z = 2t-1\)
so in standard form the line is
\( \displaystyle \frac{x-3}{2}=\frac{y}{1}=\frac{z+1}{2} \)
Its direction ratios are 2, 1, 2, and \(\sqrt{4+1+4} = 3\). Dividing by 3 gives the direction cosines:
\( \displaystyle \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\) — Incorrect. This comes from direction ratios 1, 2, 1, which turns the coefficients upside down; when y increases by 1, x and z increase by 2.
Option (2) \(\displaystyle\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\) — Incorrect. x, y and z all increase together, so the y-component cannot have the opposite sign.
Option (4) \(\displaystyle\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, -\frac{1}{\sqrt{11}}\) — Incorrect. This uses the constants 3 and −1, which fix a point of the line, not its direction.