Question

If \(l,m,n\) are direction cosines of a vector \(\bar{r}\) such that \(l+m+n=1\) and \(l^{2}+m^{2}+n^{2}=1\), then:
(where \(l,m,n\geq 0\))

• A. the vector lies along one of the coordinate axes.
• B. the vector \(\bar{r}\) is equally inclined to the three coordinate axes.
• C. The vector \(\bar{r}\) is perpendicular to the vector \(\hat{i}+\hat{j}+\hat{k}\)
• D. The vector \(\bar{r}\) makes an angle of \(\frac{\pi }{3}\) with each of the coordinate axes.

Hint:

Square the first condition and use the second to show the sum of pairwise products is zero. Non-negative values then force two cosines to be zero.

Answer 1

The Correct Option is A

Step 1: Check an Example
Take $(l,m,n)=(1,0,0)$. Both conditions hold: $1+0+0=1$ and $1+0+0=1$.

Step 2: Eliminate Other Options
For equal inclination, $l=m=n=\dfrac1{\sqrt3}$, but then $l+m+n=\sqrt3\neq1$. The angle $\dfrac\pi3$ gives $l=m=n=\dfrac12$, with squares summing to $\dfrac34\neq1$.

Step 3: Perpendicular Option
Perpendicular to $\hat i+\hat j+\hat k$ means $l+m+n=0$, which contradicts $l+m+n=1$.

Final Answer:
Only the coordinate axis option is consistent with both conditions. \[ \boxed{\text{(A) }\text{the vector lies along one of the coordinate axes.}} \]