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Three Dimensional Geometry

Short Answer Type

211. Find the equation of the plane through the points (0, – 1, 0), (2, 1, –1) and (1,1,1).

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212. Find the equation of the plane passing through the points (0. – 1, 0), (1, 1, 1) and (3, 3,0).

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Long Answer Type

213. Find the equation of the plane through the three points (1, 1, 1), (1, – 1, 1) and (–7, – 3,–5).

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Short Answer Type

214.

Find the vector equation of the plane passing through the point A(2, 2, –1), B(3, 4, 2) and C (7, 0, 6). Also find the cartesian equation of the plane.

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215. Find the vector equations of the plane passing through the points R(2, 5, –3), S(– 2, – 3, 5) and T(5, 3, – 3).

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216. Find the equations of the planes that passes through three points:
(1, 1,–1), (6, 4, –5),(–4, –2, 3)

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217. Find the equations of the planes that passes through three points:
(1, 1, 0), (1, 2, 1), (–2, 2, –1)

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Long Answer Type

218. If from a point P (a, b, c) perpendiculars PA and PB are drawn to yz and zx-planes, then find the vector equation of the plane OAB.

Point P is (a, b, c)
PA ⊥ y z-plane and PB ⊥ z x-plane.
∴ A is (0, b, c) and B is (a, 0, c)
We are to find the equation of plane through (0, 0, 0), (0, b, c) and (a, 0, c).
The equation of plane through (0, 0, 0) is
λ (x – 0) + μ (y – 0) + v (z – 0) = 0
∴ λx + μ y + v z = 0 ...(1)
∴ it passes through (0, b, c) and (a, 0, c)
∴ 0 λ + b μ + c v = 0
and a λ + 0 μ + c v = 0
Solving these, we get,
$fraction numerator straight lambda over denominator bc minus 0 end fraction space equals fraction numerator straight mu over denominator straight c space straight a minus 0 end fraction space equals space fraction numerator straight v over denominator 0 minus ab end fraction$

$therefore space space space straight lambda over bc space equals space straight mu over ca space equals space fraction numerator straight v over denominator negative ab end fraction space equals space straight k space left parenthesis say right parenthesis therefore space space space space space space space space space space straight lambda space equals space straight k space straight b space straight c comma space space space straight mu space equals space straight k space straight c space straight a comma space space space straight v space equals space minus space straight k space straight a space straight b$

Putting values of λ, μ,v in (1), we get,
k b c x + k c a y – k a b z = 0
or $straight x over straight a plus straight y over straight b minus straight z over straight c space equals space 0 comma space which space is space required space of space plane. space$

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219. Find the equation of plane through the points (2, 2, 1), (9, 3, 6), and perpendicular to the plane 2x + 6y + 6z = 9.

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220. Find the equation of the plane through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z – 1 = 0.

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