JEE Advanced 2023 · Introduction to Three Dimensional Geometry
Let P be the plane 3x+2y+3z=16 and let S={αi^+βj^+γk^:α2+β2+γ2=1 and the distance of (α,β,γ) from the plane P is 27}.
Let u,v and w be three distinct vectors in S such that ∣u−v∣=∣v−w∣=∣w−u∣.
Let V be the volume of the parallelepiped determined by vectors u,v and w.
Then the value of 380V is
Let P be the plane √3x + 2y + 3z = 16 and let S = {αî + βĵ + γk̂ : α² + β² + γ² = 1 and the distance of (α, β, γ) from the plane P is 7/2}. Let u⃗, v⃗ and w⃗ be three distinct vectors in S such that |u⃗ - v⃗| = |v⃗ - w⃗| = |w⃗ - u⃗|. Let V be the volume of the parallelepiped determined by vectors u⃗, v⃗ and w⃗. Then the value of 80/√3V is
P:3x+2y+3z=16S={αi^+βj^+γk^:α2+β2+γ2=1,d((α,β,γ),P)=27}∣u−v∣=∣v−w∣=∣w−u∣Find 380V,V=[uvw]1.d(r,P)=27⇒n⋅r2.S is a circle: centre, radius3.V=∣u⋅(v×w)∣,find 380V∣r∣=1:α2+β2+γ2=1n=(3,2,3),∣n∣=3+4+9=4d=4∣n⋅r−16∣=27∣n⋅r∣≤∣n∣∣r∣=4⇒n⋅r−16<016−n⋅r=14⇒n⋅r=2S:∣r∣=1,n⋅r=2OS=∣n∣22n=(83,41,83),∣OS∣=21ρ=1−41=23u,v,w∈S,∣u−v∣=∣v−w∣=∣w−u∣s=ρ3=23⋅3=23V=∣u⋅((v−u)×(w−u))∣(v−u)×(w−u)=2(Area)n^n^⋅u=42=21V=2(Area)⋅21=AreaArea=43s2=43⋅49V=1693380V=380⋅169345
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