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Question:

Consider a pyramid OPQRS located in the first octant (x≥0, y≥0, z≥0) with O as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with OP=3. The point S is directly above the mid-point T of diagonal OQ such that TS=3. Then

The perpendicular distance from O to the straight line containing RS is √15/2

The acute angle between OQ and OS is π/3

The equation of the plane containing the triangle OQS is x−y=0

The length of the perpendicular from P to the plane containing the triangle OQS is 3√2

Solution:

s=(3/2,3/2,3) ⇒ OQ=3^i+3^j ⇒ OS=3/2^i+3/2^j+3^k
cosθ=(1/2+1/2)/√2√1/4+1/4+1=1/√2√3/2=1/√3
θ=π/3
n=OQ × OS=(^i+^j) × (^i+^j+2k)=2^i−^j
Equation of plane: x−y=λ ⇒ Equation of the plane: x=y
Length of perpendicular from (3,0,0) on x−y=0 is 3√2
RS → x=y=z/3=λ ⇒ x=3/2λ, y=λ+3, z=3λ
Distance from (0,0,0) → √(9/4λ^2+(3/2λ)^2+9λ^2)=√27/2λ^2+9λ^2=√27/2λ^2+9λ^2 ⇒ λ=1/3