Answer:
(i)
if θ = 900 then cos 900 = 0
A1A2 + B1B2 + C1C2 = 0
By comparing with the standard equation of a plane,
A1 = 3, B1 = 4, C1 = -5
A2 = 2, B2 = 6, C2 = 6
LHS = A1A2 + B1B2 + C1C2
= (3 × 2) + (4 × 6) + (-5 × 6)
= 6 + 24 – 30
= 0
= RHS
The angle between the planes is 900.
Hence, proved.
(ii)
if θ = 900 then cos 900 = 0
A1A2 + B1B2 + C1C2 = 0
By comparing with the standard equation of a plane,
A1 = 1, B1 = -2, C1 = 4
A2 = 18, B2 = 17, C2 = 4
LHS = A1A2 + B1B2 + C1C2
= (1 × 18) + (-2 × 17) + (4 × 4)
= 18 + (-34) + 16
= 0
= RHS
The angle between the planes is 900.
Hence, proved.