Calculus 3 Planes

Question

Find the acute angle (in degrees) between the planes 3x−2y+5z=−8 and 5x−y+3z=9.

Answer 1

given

3x-2y+5z=-8

5x-y+3z=9

the planes formula for angle between them is, A₁x+B₁y+C₁z+D₁=0 and A₂x+B₂y+C₂z+D₂=0

the acute angle is, cosθ= (A₁A₂+B₁B₂+C₁C₂)/(sqrt A₁²+B₁²+C₁² *sqrtA₂²+B₂²+C₂²)

3x-2y+5z+8=0 A₁=3, B₁=-2, C₁=5, D₁=8

5x-y+3z-9=0 A₂=5, B₂=-1 ,C₂=3, D₂=-9

cosθ=(3(5)+-2(-1)+5(3))/(sqrt9+4+25* sqrt25+1+9)

cosθ=32/sqrt38*sqrt35= 32/36.469165

cosθ=0.877454

θ=28.66

the acute angle is 28.66º

Answer 2

The angle between the two intersecting planes is equal

the acute angle the two normal vectors form.

Then if we denote by φ the angle the two normal vectors n₁=< 3, -2, 5 > and n₂ = < 5, -1, 3 >

cosφ =( n₁ ⋅ n₂)/( ||n₁|| ||n₂||

cosφ = 32/(√38 √35 ) φ ≈ 28.66º