A vector \(\vec{r}\) is inclined at equal angles to the three axes. If the magnitude of \(\vec{r}\) is \(3\sqrt{3}\) units, then the value of \(\vec{r}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pm 3(\hat{\imath} + \hat{\jmath} + \hat{k})\)
Equal angles with the three axes means the direction cosines are equal: l = m = n. Since \(l^2 + m^2 + n^2 = 1\):
\(3l^2 = 1\)
\( \displaystyle l = m = n = \pm\frac{1}{\sqrt{3}} \)
A vector is its magnitude times its unit vector \(l\hat{\imath} + m\hat{\jmath} + n\hat{k}\):
\( \displaystyle \vec{r} = 3\sqrt{3}\times\left(\pm\frac{1}{\sqrt{3}}\right)(\hat{\imath} + \hat{\jmath} + \hat{k}) \)
\(\vec{r} = \pm 3(\hat{\imath} + \hat{\jmath} + \hat{k})\)
Why the other options are incorrect
Option (2) \(\pm 2(\hat{\imath} + \hat{\jmath} + \hat{k})\) — Incorrect. Equal angles, but the magnitude is \(2\sqrt{3}\), not \(3\sqrt{3}\).
Option (3) \(\pm(\hat{\imath} + \hat{\jmath} + \hat{k})\) — Incorrect. Equal angles, but the magnitude is \(\sqrt{3}\).
Option (4) \(\pm 3(\hat{\imath} - \hat{\jmath} - \hat{k})\) — Incorrect. The magnitude is \(3\sqrt{3}\), but the direction cosines are \(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\) (or their negatives): the angle with the x-axis differs from the other two.