Question

A unit radial vector $\hat{r}$ makes angles of $\alpha ={30}^o$ relative to the $x$-axis, $\beta ={60}^o$ relative to the $y$-axis, and $\gamma ={90}^o$ relative to the $z-$axis. The vector $\hat{r}$ can be written as:

• A. $\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$
• B. $\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$
• C. $\frac{\sqrt{2}}{3}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

Given:
$\alpha ={30}^{\circ }$
$\beta ={60}^{\circ }$
$\gamma ={90}^{\circ }$

We know:
$\cos \alpha =\frac{A_x}{\left|r\right|}$
$\cos \beta =\frac{A_y}{\left|r\right|}$
$\cos \gamma =\frac{A_z}{\left|r\right|}$

where, $A_x,A_y,A_z$ are the co-ordinates of unit radial vector along respective directions.

Hence,
$A_x=\cos \alpha \left|r\right|$
$\Rightarrow A_x=\cos {30}^{\circ }\left(1\right)=\frac{\sqrt{3}}{2}$
$A_y=\cos \beta \left|r\right|$
$\Rightarrow A_y=\cos {60}^{\circ }\left(1\right)=\frac{1}{2}$
$A_z=\cos \gamma \left|r\right|$
$\Rightarrow A_z=\cos {90}^{\circ }\left(1\right)=0$

Hence, $r=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}+0\hat{k}=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$