Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equations of planes are $x+y+2z-6=0$ and $2x-y+z-9=0$ $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$ $=\frac{1\times 2+1\times (-1)+2\times 1}{\sqrt{1+1+4}\sqrt{4+1+1}}$ $=\frac{2-1+2}{\sqrt{6}\times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$ $\Rightarrow \cos \theta =\cos \left(\frac{\pi }{3}\right)$ $\therefore \theta =\frac{\pi }{3}$