Consider the plane E : 
= 
+ λ 
+ μ 
F is a plane containing the point A (–4, 2, 2) and parallel to E. Suppose the point B is on the plane E, such that B has a minimum distance from point A. If C(–3, 0, 4) lies in the plane F. Then find the area of Δ ABC.
Text Solution
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Sol. E Planes
= 
= 
2 (x + 1) – (y – 1) – 2(z – 1) = 0
2x – y – 2z + 5 = 0
Plane F : 2x – y – 2z + k = 0 pass through (–4, 2, 2)
–8 – 2 – 4 + k = 0 ⇒ k = 14
2x – y – 2z + 14 = 0
A(–4, 2, 2) B is the foot of ⊥ from A on the plane E
E : (2x – y – 2z + 5 = 0)
= 
= 
= – 
(x, y, z) = (–2, 1, 0) = B
C = (–3, 0, 4)
area of Δ ABC = 
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