A vector \(\vec{r}\) of magnitude 6 makes an angle of \(\displaystyle\frac{\pi}{3}\) with \(\hat{\imath}\), an angle of \(\displaystyle\frac{2\pi}{3}\) with \(\hat{\jmath}\) and an obtuse angle with \(\hat{k}\). Then \(\vec{r}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3\hat{\imath}-3\hat{\jmath}-3\sqrt{2}\,\hat{k}\)
The direction cosines of \(\vec{r}\) are
\( \displaystyle l = \cos\frac{\pi}{3} = \frac{1}{2} \)
\( \displaystyle m = \cos\frac{2\pi}{3} = -\frac{1}{2} \)
From \(l^2+m^2+n^2 = 1\):
\( \displaystyle n^2 = 1-\frac{1}{4}-\frac{1}{4} = \frac{1}{2} \)
The angle with \(\hat{k}\) is obtuse, so n is negative:
\( \displaystyle n = -\frac{1}{\sqrt{2}} \)
Then \(\vec{r} = 6(l\hat{\imath}+m\hat{\jmath}+n\hat{k})\):
\( \displaystyle \vec{r} = 6\times\frac{1}{2}\hat{\imath}-6\times\frac{1}{2}\hat{\jmath}-6\times\frac{1}{\sqrt{2}}\hat{k} \)
\( = 3\hat{\imath}-3\hat{\jmath}-3\sqrt{2}\,\hat{k}\)
Why the other options are incorrect
Option (1) \(3\hat{\imath}-3\hat{\jmath}+3\sqrt{2}\,\hat{k}\) — Incorrect. Its z-component is positive, so the angle with \(\hat{k}\) is acute.
Option (3) \(3\hat{\imath}+3\hat{\jmath}-3\sqrt{2}\,\hat{k}\) — Incorrect. Its y-component is positive, so the angle with \(\hat{\jmath}\) would be \(\frac{\pi}{3}\), not \(\frac{2\pi}{3}\).
Option (4) \(6\hat{\imath}-6\hat{\jmath}-6\sqrt{2}\,\hat{k}\) — Incorrect. Its magnitude is \(\sqrt{36+36+72} = 12\), not 6.