Question

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

$7x-3y+5z=6$ and $2x+3y-z=1$

Short answer

The angle between two planes is $\theta =\frac{\pi }{2}$

Step-by-step solution

Given information

The two planes $7x-3y+5z=6$and $2x+3y-z=1$

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 $7x-3y+5z=6$ is ${\mathbf{N}}_1=⟨7,-3,5⟩$ and the normal vector of the plane $2x+3y-z=1$ is ${\mathbf{N}}_2=⟨2,3,-1⟩$

The angle between the two planes is:

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

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

$={\cos }^{-1}\left(\frac{14-9-5}{\sqrt{83}\sqrt{14}}\right)$ (Simplify)

$={\cos }^{-1}\left(\frac{0}{\sqrt{83}\sqrt{14}}\right)$

$={\cos }^{-1}\left(0\right)$

$=\frac{\pi }{2}$(Rationalize)

The angle between two planes is $\theta =\frac{\pi }{2}$