Question

Find the angle between the planes $x-2y+2z-3=03,x+4y+12z+1=0$

Answer 1

Consider the given equation of the plane .

$A=x-2y+2z-3=0,B=x+4y+12+1=0$

Let, Angle between given two planes =$\theta$

$\cos \theta =$ $\cos \theta =\frac{A.B}{\left|A\right|\left|B\right|}=$

$=\frac{1.1+(-2)\left(4\right)+2.12}{\sqrt{1^2+{(-2)}^2+{\left(2\right)}^2}\sqrt{1^2+4^2+{12}^2}}$

$=\frac{1-8+24}{3.\sqrt{161}}=\frac{17}{3.\sqrt{161}}$

$\theta ={\cos }^{-1}\left(\frac{17}{3.\sqrt{161}}\right)$

Hence this is the answer ,