# unable to find angle of plane $7x+13y+4z = 9$ with $X-Y$ plane

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

$ax+by+0.z=d$ for $(X-Y)$ palne

Thanks

-
The X-Y plane usually denotes the plane spanned by both the X-axis and the Y-axis, that is, $0x+0y+1z=0$. –  DoomMuffins Jan 25 '13 at 14:30

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.

-

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

-
And finally, you can compute an approximation of this value. But a) you shouldn't add your computations as an answer to your question (that's why someone downvoted it) b) you should upvote/accept useful answers ;) –  Klaus Jan 25 '13 at 19:52