Which of the following can be the direction cosines of a vector?

The correct answer is Option (1) → \(\displaystyle\frac{1}{\sqrt{2}}, -\frac{1}{2}, \frac{1}{2}\)

Direction cosines l, m, n of any vector always satisfy the basic relation

    \(l^2+m^2+n^2 = 1\)

For Option (1):

    \( \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 1 \)

So these can be direction cosines, for example of the vector \(\sqrt{2}\,\hat{\imath}-\hat{\jmath}+\hat{k}\), whose magnitude is 2.

Why the other options are incorrect

• Option (2) \(\displaystyle\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}\) — Incorrect. The sum of squares is \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4} = \frac{3}{4}\).

• Option (3) \(\displaystyle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{3}\) — Incorrect. The sum of squares is \(\frac{1}{3}+\frac{1}{3}+\frac{1}{9} = \frac{7}{9}\).

• Option (4) \(\displaystyle\frac{2}{3}, -\frac{1}{3}, \frac{1}{3}\) — Incorrect. The sum of squares is \(\frac{4}{9}+\frac{1}{9}+\frac{1}{9} = \frac{2}{3}\).