The direction cosines of the vector \(\overrightarrow{AB}\), where A and B are the points (1, −2, 3) and (−1, 2, −1) respectively, are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle-\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\)
The vector \(\overrightarrow{AB}\) is the position vector of B minus the position vector of A:
\(\overrightarrow{AB} = (-1-1)\hat{\imath}+(2+2)\hat{\jmath}+(-1-3)\hat{k}\)
\( = -2\hat{\imath}+4\hat{\jmath}-4\hat{k}\)
Its magnitude is
\(|\overrightarrow{AB}| = \sqrt{4+16+16} = 6\)
The direction cosines are the components divided by the magnitude:
\( \displaystyle \frac{-2}{6}, \frac{4}{6}, \frac{-4}{6} \)
\( \displaystyle = -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\) — Incorrect. These are the direction cosines of \(\overrightarrow{BA}\), the opposite vector.
Option (2) −2, 4, −4 — Incorrect. These are direction ratios; their squares add up to 36, not 1.
Option (3) \(\displaystyle-\frac{1}{5}, \frac{2}{5}, -\frac{2}{5}\) — Incorrect. The components are divided by 10, the sum of their sizes, instead of by the magnitude 6.