What is the angle between following planes $2x-3y+4z=1$ and $-x+y=4$?

Answer 1

Toolbox:

• The angle between two planes is the angle between the normals to the two planes. $ \theta=\cos^{-1}\bigg( \large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|} \bigg).$ If $ \overrightarrow n_1.\overrightarrow n_2=0$ the planes are at right angles.

$ 2x-3y+4z=1$ (i) and $-x+y=4$ (ii)

$ \overrightarrow n_1 = 2\overrightarrow i-3\overrightarrow j+4\overrightarrow k \: \: \overrightarrow n_2=-\overrightarrow i+\overrightarrow j$

$ \theta = \cos^{-1} \large\frac{[\overrightarrow n_1.\overrightarrow n_2]}{|\overrightarrow n_1||\overrightarrow n_2|}$$=\cos^{-1} \bigg[ \large\frac{(2)(-1)+(-3)(1)+0}{\sqrt{4+9+16}\sqrt{1+1}} \bigg]$

$ = \cos^{-1} \large\frac{-5}{\sqrt{58}}$