Find the angle between the planes: 1^° x+2z-6=0 and x+2y-4=0…

Find the angle between the planes: $1^\circ$ $x+2z-6=0$ and $x+2y-4=0$; $2^\circ$ $\vec r\cdot(\vec i-2\vec j+2\vec k)=8$ and $\vec r\cdot(\vec i+\vec k)=6$.

Explanation of Method

Angles between two pairs of planes, computed from their normal vectors.

Step 1: Formula

\[ \cos\varphi=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1|\,|\vec n_2|}\qquad(0\le\varphi\le90^\circ) \]

Step 2: $1^\circ$

$\vec n_1=\{1,0,2\}$ and $\vec n_2=\{1,2,0\}$, so $\vec n_1\cdot\vec n_2=1$ and $|\vec n_1|=|\vec n_2|=\sqrt5$.

\[\cos\varphi=\frac15,\qquad \varphi=\arccos\frac15\approx78.46^\circ\approx78^\circ28'\]

Step 3: $2^\circ$

$\vec n_1=\{1,-2,2\}$ and $\vec n_2=\{1,0,1\}$, so $\vec n_1\cdot\vec n_2=3$, $|\vec n_1|=3$ and $|\vec n_2|=\sqrt2$.

\[\cos\varphi=\frac{3}{3\sqrt2}=\frac{1}{\sqrt2},\qquad \varphi=45^\circ\]

$\varphi$ $\varphi$ $\pi_1$ $\pi_2$ $\vec n_1$ $\vec n_2$

Cross-section perpendicular to the line of intersection: the angle between the planes equals the angle between their normals (or its supplement), hence $\cos\varphi=\tfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$.

Result

\[\boxed{\,\cos\varphi=\frac15,\qquad \varphi=\arccos\frac15\approx78.46^\circ\approx78^\circ28'\,}\]

\[\boxed{\,\cos\varphi=\frac{3}{3\sqrt2}=\frac{1}{\sqrt2},\qquad \varphi=45^\circ\,}\]