Question

What are the direction cosines of $\overrightarrow{r}$ ?

• A. $(\frac{1}{2},-\frac{\sqrt{3}}{2},0)$
• B. $(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
• C. $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
• D. $(-\frac{1}{2},\frac{\sqrt{3}}{2},0)$

Answer 1

Answer: (B)

$(\frac{1}{2},\frac{\sqrt{3}}{2},0)$

Sum of squares of direction cosines $=1$

$l^2+m^2+n^2=1$

Here, $l=\cos {60}^0,m=\cos {30}^0,n=?$

$n^2=1-l^2-m^2{n}^2=1-{\left(\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2=0$

Hence, $n=0=\cos \theta$

$\Rightarrow \theta ={90}^0$

Direction cosines $(l,m,n)\equiv (\frac{1}{2},\frac{\sqrt{3}}{2},0)$