Determine the angles formed by the given line with the coordinates axis $ (D): { x-2y-5=0} / {x-3z+8=0 } $ ( this is a system of 2 equations, I don't know how to format it properly )
I know the directionals are ( 1, -2 , 0) and (1,0,-3) but I don't really know how to continue it.
just a hint to finish
The line $(D) $ can be defined by following parametric equations
$$x=x $$ $$y=(x-5)/2$$ $$z=(x+8)/3$$
thus the vector director of $(D) $ is $$\vec {u}=(1,1/2,1/3) $$ or $$(6,3,2) $$
use $$\cos (\vec{u},\vec{v})=\frac {\vec {u}\cdot \vec{v}}{||\vec {u}||||\vec {v}||} $$
The system of equations describes the line as the intersection of two planes. It must therefore be perpendicular to the normals of both planes, so its direction vector is $(1,0,-3)\times(1,-2,0)=(6,-3,2)$. The corresponding unit vector is $\left(\frac67,-\frac37,\frac27\right)$. Since $v\cdot w = \|v\|\,\|w\|\cos\theta$, the components of this vector are the cosines of the angles it makes with the coordinate axes.
