The direction cosines l, m, n of a line satisfy \(3l = 2m = n\). Which of the following can be the direction cosines of the line?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\)
Let the common value be t, so \(3l = 2m = n = t\):
\( \displaystyle l = \frac{t}{3},\quad m = \frac{t}{2},\quad n = t \)
Multiplying by 6, the direction ratios are
\(l : m : n = 2 : 3 : 6\)
Dividing by \(\sqrt{2^2+3^2+6^2} = \sqrt{49} = 7\) gives the direction cosines:
\( \displaystyle \pm\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right) \)
Check: \(3\cdot\frac{2}{7} = 2\cdot\frac{3}{7} = \frac{6}{7}\).
Why the other options are incorrect
Option (2) \(\displaystyle\frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{1}{\sqrt{14}}\) — Incorrect. This takes the coefficients 3, 2, 1 as direction ratios; then \(3l = \frac{9}{\sqrt{14}}\) but \(2m = \frac{4}{\sqrt{14}}\).
Option (3) \(\displaystyle\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\) — Incorrect. Here \(3l = \frac{18}{7}\) but \(2m = \frac{6}{7}\).
Option (4) \(\displaystyle\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}\) — Incorrect. Here \(3l = \frac{6}{7}\) but \(2m = -\frac{6}{7}\).