Question

Find the angle between the planes x+2y+2z-5=0 and 3x+3y+2z-8=0.

Answer 1

$P_1:x+2y+2z-5=0,P_2:3x+3y+2z-8=0$

$a_1=1,b_1=2,c_1=2a_2=3,b_2=3,c_2=2$

$\cos \theta =\frac{a_1{a}_2+b_1+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}=\frac{1\times 3+2\times 3+2\times 2}{\sqrt{1^2+2^2+2^2}\sqrt{3^2+3^2+2^2}}$

$=\frac{3+6+4}{\sqrt{9}\sqrt{22}}=\frac{13}{3\sqrt{22}}$

$\therefore \theta ={\cos }^{-1}\frac{13}{3\sqrt{2^2}}$