Calculate the angle at which the planes intersect

[mati89]

Calculate the angle at which the planes intersect
-x-2y+z-5=0 i -2x+y-2z+1=0

[ognjenmi]

Angle between two planes is equal to the angle between straight lines normal to those planes. Such straight line is represented by vector [tex]\vec{n} = A\vec{i}+B\vec{j}+C\vec{k}[/tex] where A, B and C are coefficients in plane equation [tex]Ax + By +Cz + D = 0[/tex].

Knowing that scalar product of two vectors is [tex]\vec{a}\cdot\vec{b} = \left | \vec{a} \right |\left | \vec{b} \right |cos\alpha[/tex], where [tex]\alpha[/tex] is angle between those vectors we can easily calculate the angle between two planes.

Plane 1: [tex]A_1=-1, B_1=-2, C_1=1; \vec{n_1}=-1\cdot\vec{i}-2\cdot\vec{j}+1\cdot\vec{k}[/tex]
Plane 2: [tex]A_2=-2, B_2=1, C_2=-2; \vec{n_2}=-2\cdot\vec{i}+1\cdot\vec{j}-2\cdot\vec{k}[/tex]

And therfore:
[tex]cos\alpha = \frac{\vec{n_1}\cdot\vec{n_2}}{\left| \vec{n_1} \right| \left|\vec{n_2}\right|}[/tex]
[tex]cos\alpha = \frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^{2} + B_1^{2} + C_1^{2}}\sqrt{A_2^{2} + B_2^{2} + C_2^{2}} }=\frac{-2}{\sqrt{6}\cdot\sqrt{9}}=-\frac{\sqrt{6}}{9}[/tex]

[tex]\alpha=arccos(-\frac{\sqrt{6}}{9})\approx0.5877\pi\approx105.79^\circ[/tex]