Answer:
a) P.V of is OP₁ = [ 1.5i + 0j + 2.6k ], P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]
b) Vector connecting P₁ to P₂ is [ 0i + 1.5j + 0.48k ]
c) cylindrical coordinates are (1.5, π/2, 0.48)
Step-by-step explanation:
Given that;
r = 3
P₁ ( 3, 0°, 30° ), P₂ ( 3, 45°, 45° )
a)
P.V of P₁
x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5
y = rsin∅sin∅ = 3(sin0°) (sin30°) = (3 × 0 × 0.5) = 0
z = rcos∅ = 3(cos30°) = ( 3 × 0.866) = 2.6
∴ P.V of is OP₁ = [ 1.5i + 0j + 2.6k ]
P.V of P₂
x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5
y = rsin∅sin∅ = 3(sin45°) (sin45°) = (3 × 0.7071 × 0.7071) = 1.5
z = rcos∅ = 3(cos45°) = ( 3 × 0.7071) = 2.12
∴ P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]
b)
Vector connecting P₁ to P₂ is given by
OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]
= [ 0i + 1.5j + 0.48k ]
c)
P₁P₂ → = [ 0i + 1.5j + 0.48k ] = [ 1.5j + 0.48k ]
so in a cylindrical coordinate, it should be
r = √(o² + 1.5²) = 1.5
∅ = tan⁻¹[y/π] = π/2
z = 0.48
cylindrical coordinates are (1.5, π/2, 0.48)
