How to find angle of plane $7x+13y+4z = 9$ with $xy$ coordinate plane?

Question

How can I calculate inclination of $7x+13y+4z = 9$ with $X-Y$ plane

As for as I understand from question is that the angle of plane $7x+13y+4z=9$ with $ax+by+0z=d$ for $(XY)$ plane.

Answer 1

As you probably know, a vector perpendicular to a plane $$ax+by+cz=d$$ is given by $(a,b,c)$.

Now, the angle between two planes is precisely the acute angle between two normal vectors. Using the formula $$\cos{\angle (\bf{v},\bf{w}})=\frac{|\bf{v}\cdot\bf{w}|}{||\bf{v}||||\bf{w}||},$$ you can easily perform your calculation.

Answer 2

Thanks DoomMuffins and Klause.

So our equation of plane is $7x+13y+4z=9$ and $0.x+0.y+0.z=0$

So Normal vectors is $<7,13,4>$ and $<0,0,1>$

So $\displaystyle \cos \theta = \frac{<7,13,4>.<0,0,1>}{\sqrt{7^2+13^2+4^2}\sqrt{1}}=\frac{4}{\sqrt{234}}$

So $\displaystyle \theta = \cos^{-1}\left(\frac{4}{\sqrt{234}}\right)$