Let \(l\), \(m\) and \(n\) be the direction cosines of a non-zero vector \(\vec{r}\). Which of the following statements are correct?
(A) \(l^2+m^2+n^2 = 1\)
(B) \(l+m+n = 1\)
(C) \(l\hat{\imath}+m\hat{\jmath}+n\hat{k}\) is the unit vector in the direction of \(\vec{r}\)
(D) \(-l\), \(-m\) and \(-n\) are the direction cosines of \(-\vec{r}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Let \(\vec{r} = x\hat{\imath}+y\hat{\jmath}+z\hat{k}\) and \(r = |\vec{r}|\). The direction cosines are
\( \displaystyle l = \frac{x}{r}, \quad m = \frac{y}{r}, \quad n = \frac{z}{r} \)
\( \displaystyle l^2+m^2+n^2 = \frac{x^2+y^2+z^2}{r^2} = 1 \)
So the squares of \(l, m, n\) add up to 1, the vector \(l\hat{\imath}+m\hat{\jmath}+n\hat{k}\) is \(\frac{\vec{r}}{r}\), and reversing \(\vec{r}\) changes the sign of every component.
Checking the statements
(A) \(l^2+m^2+n^2 = 1\) — Correct. \(x^2+y^2+z^2 = r^2\).
(B) \(l+m+n = 1\) — Incorrect. For \(\vec{r} = \hat{\imath}+\hat{\jmath}+\hat{k}\), \(l+m+n = \frac{3}{\sqrt{3}} = \sqrt{3}\).
(C) \(l\hat{\imath}+m\hat{\jmath}+n\hat{k}\) is the unit vector along \(\vec{r}\) — Correct. It equals \(\frac{\vec{r}}{r}\), whose length is 1.
(D) \(-l, -m, -n\) are the direction cosines of \(-\vec{r}\) — Correct. \(-\vec{r}\) has components \(-x, -y, -z\) and the same length \(r\).