The direction cosines of a line which lies in ZoX plane and makes an angle of $30^{\circ}$ with Z-axis are
The direction cosines of a line which lies in ZoX plane and makes an angle of $30^{\circ}$
with Z-axis are
$0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}$
$\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}$
$0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}$
$\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}$
Solution
Direction cosines are the cosines of the angles which a line makes with the positive coordinate axes in anticlockwise direction. They are represented by $\langle l, m, n\rangle$ where $\mathrm{l}, \mathrm{m}, \mathrm{n}$ correspond to the $\mathrm{x}$-axis, $\mathrm{y}$-axis, and $\mathrm{z}$-axis respectively. One important property of direction cosines is that the sum of their squares is unity.
i.e., $l^{2}+m^{2}+n^{2}=1$
If the line lies in the zox plane, it implies that the $y$-axis is normal to the line as the $y$-axis is normal to the ZOX plane. Which means the angle the line makes with the $y$-axis is $90^{\circ}$ or $\frac{\pi}{2}$ radians. Thus, $m=\cos \left(\frac{\pi}{2}\right)=0$
It is given that the line makes an angle of $30^{\circ}$ or $\frac{\pi}{6}$ radians with the positive z-axis. Thus, $n=\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$
Now we use the property $l^{2}+m^{2}+n^{2}=1$
Substituting the values of $m$ and $n$, we get the following values of । $l=\pm \frac{1}{2}$
We get two values of I because the only information given is that the line makes an angle of $\frac{\pi}{6}$ with the z-axis, so it can make an angle of either $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$ with the $x$-axis, which correspond to the two values of I.
Therefore, the direction cosines are
$\langle l, m, n\rangle=\left\langle\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right\rangle$