Find the acute angle (in degrees) between the planes -2z-5y-z and 4y+2z=8

Answer 1

First, find the dot product of both functions (assuming the first equation is supposed to read -2x-5y-z=0 and the second one 4y+2z=8, also note that the second function can be written as 2y+z=4):

<-2,-5,-1> • <0,2,1> = (-2)(0) + (-5)(2) + (-1)(1) = -10 - 1 = -11

Next, we find the magnitude of the two planes and find their product:

|<2,-5,-1>| = √((2)² + (-5)² + (-1)²) = √(4+25+1) = √30

|<0,2,1>| = √(0² + 2² + 1²) = √(4+1) = √5

√30•√5 = √6√5•√5 = 5√6

Now take our dot product of -22 from earlier and divide it by our magnitude product of 10√6 while rationalizing the denominator:

-11/(5√6) = (-11√6)/(5*6) = (-11√6)/30

Lastly, find the inverse cosine of the result:

cos^-1((-11√6)/30) ≈ 153.915°

As this is not an acute angle, we simply find its complementary:

α = 180° - θ

α = 180° - 153.915°

α ≈ 26.085°

Thus, the acute angle between the two planes is about 26.085°

Hope this helps! Let me know if the equations for the original planes were not correct and I can fix my answer.