Note: Below solutions were found by me while learning the material. They are unchecked and many are likely incorrect. Final answers are underlined.
a) A,B,C,D = (+1,+1,+1), (+1,-1,-1), (-1,+1,-1), (-1,-1,+1)
b) A ⋅ B = A B cos(θ) = -1 ⟹ θ = acos(-1/3) ≈ 109°
c)
- AB ⋅ AC = 8 cos(θ) = 4 ⟹ θ = acos(1/2) = π/3 = 60°
- AB ⋅ CD = 8 cos(θ) = 0 ⟹ θ = acos(0) = π/2 = 90°
d) (1/2) * 2√2 * 2√2 * sin(π/3) = 2√2
a) v₁ ⋅ (v₂-v₃) = 0
b)
- P lies on the altitude from P₁ ⟺ v₁ ⋅ (v₂-v₃) = 0 ⟺ v₁⋅v₂ - v₁v₃ = 0 ⟺ v₁⋅v₂ = v₁v₃
- P lies on the altitude from P₂ ⟺ v₂ ⋅ (v₁-v₃) = 0 ⟺ v₁⋅v₂ - v₂v₃ = 0 ⟺ v₁⋅v₂ = v₂v₃
c) v₁v₃ = v₂v₃ ⟺ v₁v₃ - v₂v₃ = 0 ⟺ v₃ ⋅ (v₁-v₂) ⟺ P lies on the altitude from P₃
The tetrahedron has rotational symmetry in 3 dimensions, so the vectors must also have rotational symmetry in 3 dimensions. Any symmetrical set of vectors sums to 0.
a) u: /, v: \
b) For clarity, let a=θ_1 and b=θ_2
A₁A₂ =
cos(a)cos(b)-sin(a)sin(b) -cos(a)sin(b)-sin(a)cos(b)
sin(a)cos(b)+sin(b)cos(a) -sin(a)sin(b)+cos(a)cos(b)
=
cos(a+b) -sin(a+b)
sin(a+b) cos(a+b)
A rotation of a degrees followed by a rotation of b degrees is the same as a rotation of a+b degrees.
c)
A^-1 = 1/(cos²(θ)+sin²(θ)) ⋅
cos(θ) sin(θ)
-sin(θ) cos(θ)
= A^T
cos²(θ)+sin²(θ) = 1
A^T⋅A =
cos²(θ)+sin²(θ) sin(θ)cos(θ)-sin(θ)cos(θ)
sin(θ)cos(θ)-sin(θ)cos(θ) cos²(θ)+sin²(θ)
=
1 0
0 1
sin(-θ) = -sin(θ), cos(-θ) = cos(θ). Rotation by -x is rotation in the opposite direction.
d)
-1/√2 √(1-1/√2) -1/√2 √(1+1/√2) -1/√2 √(1+1/√2) -1/√2 √(1-1/√2)
√(1-1/√2) 1 -√(1+1/√2) 1 √(1+1/√2) -1 -√(1-1/√2) -1
e)
a) The amount of each ingredient required to make all of the pastries.
b) N⋅M⋅X
c) No access to MATLAB
d)