Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are ${\overrightarrow{\mathrm{N}}}_1=2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ and ${\overrightarrow{\mathrm{N}}}_2=3\hat{\mathrm{i}}-6\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ Therefore, $\cos \theta =\left|\frac{{\overrightarrow{\mathrm{N}}}_1\cdot {\overrightarrow{\mathrm{N}}}_2}{\left|{\overrightarrow{\mathrm{N}}}_1\right|\left|{\overrightarrow{\mathrm{N}}}_2\right|}\right|$ $$=\left|\frac{(2\hat{i}+\hat{j}-2\hat{k})\cdot (3\hat{i}-6\hat{j}-2\hat{k})}{\sqrt{4+1+4}\sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)$$ Hence, $\theta ={\cos }^{-1}\left(\frac{4}{21}\right)$