Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given $l+m+n=1$

Squaring both sides:

$(l+m+n{)}^2=1$

$l^2+m^2+n^2+2(lm+mn+nl)=1$

Since $l^2+m^2+n^2=1$ for direction cosines:

$1+2(lm+mn+nl)=1$

$lm+mn+nl=0$

Given $l,m,n\ge 0$, the sum of non-negative terms can be zero only if at least two of $l,m,n$ are zero.

If $m=0$ and $n=0$, then $l=1$.

Thus, the possible direction cosines are $(1,0,0)$, $(0,1,0)$, or $(0,0,1)$.

This implies that the vector $\bar{r}$ lies along one of the coordinate axes.

Answer: the vector lies along one of the coordinate axes.