Question

Use your answers from Exercise 14 to find the angle between the indicated planes in Exercises 44 and 45.

$-x+7y-2z=5$ and $3x+5y-4z=2$

Short answer

The angle between two planes is $\theta ={\cos }^{-1}\left(\frac{4\sqrt{3}}{9}\right)$

Step-by-step solution

Given information

The two planes $-x+7y-2z=5$and $3x+5y-4z=2$

Calculation

The goal is to determine the angle between the two planes shown.

The angle formed by two planes can be calculated as follows:

$\theta ={\cos }^{-1}\left(\frac{{\mathbf{N}}_1·{\mathbf{N}}_2}{||{\mathbf{N}}_1||||{\mathbf{N}}_2||}\right)$

Calculation

The normal vector of the plane $-x+7y-2z=5$ is ${\mathbf{N}}_1=⟨-1,7,-2⟩$ and the normal vector of the plane $3x+5y-4z=2$ is ${\mathbf{N}}_2=⟨3,5,-4⟩$

The angle between the two planes is:

$\theta ={\cos }^{-1}\left(\frac{⟨-1,7,-2⟩·⟨3,5,-4⟩}{\Vert (-1,7,-2⟩\Vert \Vert (3,5,-4⟩\Vert }\right)$ (Substitution)

$={\cos }^{-1}\left(\frac{-1\left(3\right)+7\left(5\right)-2(-4)}{\sqrt{1+49+4}\sqrt{9+25+16}}\right)\text{(Dot Product)}$

$={\cos }^{-1}\left(\frac{40}{\sqrt{54}\sqrt{50}}\right)$ (Simplify)

$={\cos }^{-1}\left(\frac{40}{3\sqrt{6}\times 5\sqrt{2}}\right)$

$={\cos }^{-1}\left(\frac{4}{3\sqrt{3}}\right)$

$={\cos }^{-1}\left(\frac{4\sqrt{3}}{9}\right)$(Rationalize)

The angle between two planes is $\theta ={\cos }^{-1}\left(\frac{4\sqrt{3}}{9}\right)$