If a unit vector has direction cosines $l=\frac{1}{\sqrt{3}}$, $m=\frac{1}{\sqrt{3}}$, and $n=\frac{1}{\sqrt{3}}$, what is the magnitude of the vector?

Solution

1. The direction cosines $l,m,n$ of a vector satisfy $l^2+m^2+n^2=1$ if the vector is a unit vector.
2. Calculate the sum of squares: $$\begin{array}{rl}(\frac{1}{\sqrt{3}}{)}^2+(\frac{1}{\sqrt{3}}{)}^2+(\frac{1}{\sqrt{3}}{)}^2& =\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\\ & =1\end{array}$$.
3. Since the sum of squares is 1, the magnitude of the vector is 1.
4. Hence the answer is (A).