If a line makes 60^o and 45^o angles with the positive directions of  x-axis and z-axis  respectively, then find the angle that it makes with the positive direction of y-axis.  Hence, write the direction cosines of the line.

Or

Check whether the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}\text{ and }\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{z}$ are skew or not.

Given: 𝛼 = 60^𝑜 and 𝛾 = 45⁰: 
Let angles $\alpha$ makes with x-axis, $\gamma$makes with z-axis and $\beta$-makes with y-axis then 
$l=\cos {60}^{\circ },\mathrm{m}=\cos \mathrm{\beta },n=\cos {45}^{\circ }$
We know that ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2=1$
or ${\cos }^2{60}^{\circ }+{\cos }^2\mathrm{\beta }+{\cos }^2{45}^{\circ }=1$
or $\frac{1}{4}+{\cos }^2\mathrm{\beta }+\frac{1}{2}=1$
or $\begin{array}{l}{\cos }^2\mathrm{\beta }=1-\frac{1}{4}-\frac{1}{2}\\ =\frac{4-1-2}{4}\\ =\frac{1}{4}\end{array}$
or $\cos \mathrm{\beta }=\pm \frac{1}{2}$
$\therefore \mathrm{\beta }=\frac{\mathrm{\pi }}{3}\text{ or }\frac{2\mathrm{\pi }}{3}$
From the data two lines lieing in  the the interval of $\left(0,\frac{\mathrm{\pi }}{2}\right)$
Therefore, the angle that it makes with the positive direction of y-axis is $\mathrm{\beta }=\frac{\mathrm{\pi }}{3}={60}^{\circ }$
Now, for direction cossine we will use the following formula, 
If, l, m, n are the direction cosines, so, 
$\begin{array}{l}\mathrm{l}=\cos \mathrm{\alpha }=\cos {60}^{\circ }=\frac{1}{2}\\ \mathrm{m}=\cos \mathrm{\beta }=\cos {60}^{\circ }=\frac{1}{2}\\ \mathrm{n}=\cos \mathrm{\gamma }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}\end{array}$
Therefore, the angle that it makes with the positive direction of y-axis is $\mathrm{\beta }={60}^{\circ }$ and the 3 direction cosines of the line is $\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$.

Or

Given: $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-3}{4}\dots \dots \left(1\right)$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-0}{1}........\left(2\right)$
So, ${\mathrm{x}}_1=1,{\mathrm{y}}_1=2,{\mathrm{z}}_1=3$ and ${\mathrm{x}}_2=4,{\mathrm{y}}_2=1,{\mathrm{z}}_2=0$ also ${\mathrm{l}}_1=2,{\mathrm{m}}_1=3,{\mathrm{n}}_1=4$ and ${\mathrm{l}}_2=5,{\mathrm{m}}_2=2,{\mathrm{n}}_2=1$
Now for to check the lines are skew then it should not be coplanar,
So, for coplanarity,
$\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{l}}_1& {\mathrm{m}}_1& {\mathrm{n}}_1\\ {\mathrm{l}}_2& {\mathrm{m}}_2& {\mathrm{n}}_2\end{array}\right|=0$
Now, 
$\begin{array}{l}\left|\begin{array}{ccc}{\mathrm{x}}_2-{\mathrm{x}}_1& {\mathrm{y}}_2-{\mathrm{y}}_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{l}}_1& {\mathrm{m}}_1& {\mathrm{n}}_1\\ {\mathrm{l}}_2& {\mathrm{m}}_2& {\mathrm{n}}_2\end{array}\right|=\left|\begin{array}{ccc}4-1& 1-2& 0-3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|\\ \Rightarrow \left|\begin{array}{ccc}3& -1& -3\\ 2& 3& 4\\ 5& 2& 1\end{array}\right|\\ =3(3-8)-(-1)(2-20)-3(4-15)\\ =3(-5)-18-3(-11)\\ =-15-18+33=-33+33=0\end{array}$
Therefore, we can say that lines are coplanar. Hence, given lines are not skew.