Find the angle between the intersection of the two planes defined by $2x+y-z$ and $x+y+2z$ and the positive direction of $x$ axis .
I know to find the angle between two planes we just need to find the angle between their normals,first I computed the intersection of the two planes: $$-2x+z=-x+2z$$ $$\iff$$ $$z=-x$$ But I don't know what to do next
On
So, the given question asks us to find the angle made by the line with $x$-axis formed by intersection of two planes.
We know that, the line will be oriented parallel to both the planes and hence perpendicular to normal vectors of both the planes. So, the direction vector of the line can be given as $$\vec x=\vec n_1×\vec n_2$$ where $n_1$ and $n_2$ are the normal vectors of the two planes. The angle between this line and $x$-axis is $$\theta=\cos ^{-1}\left(\frac{\vec x. \hat i}{|\vec x|}\right).$$
We’ll compute the equation of the line of intersection of the planes first. As you have derived, we have $$x=-z$$ Putting this value of $x$ in one of the plane equations, $$-z+y+2z=0 \implies y=-z$$ This gives the equation of the intersection: $$\frac x1=\frac y1=\frac{z}{-1}$$ This line is directed along $(1,1-1)$ and hence makes an angle $\theta$ with $(1,0,0)$, where $$\cos\theta = \frac{(1,1-1)\cdot(1,0,0)}{|(1,1,-1)|} = \frac{1}{\sqrt 3}$$