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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.

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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.

The equation of any plane through (2, 2, 1) is
a (x – 2) + b (y – 2) + c (z – 1) = 0    ....(1)
∴ it passes through (9, 3, 6)
∴ a (9 – 2) + b (3 – 2) + c (6 – 1) = 0
∴ 7 a + b + 5 c = 0    ...(2)
Also plane (1) is perpendicular to the plane 2 x + 6 y + 6 z = 9
∴ a (2) + b (6) + c (6) = 0    [∴ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0]
∴ 2 a + 6 b + 6 c = 0
⇒    a + 3 b + 3 c = 0    ....(3)
From (2) and (3), we get, $fraction numerator straight a over denominator 3 minus 15 end fraction space equals space fraction numerator straight b over denominator 5 minus 21 end fraction space equals space fraction numerator straight c over denominator 21 minus 1 end fraction$
$therefore$ $fraction numerator straight a over denominator negative 12 end fraction space equals fraction numerator straight b over denominator negative 16 end fraction space equals space straight c over 20 space space space space space space rightwards double arrow space space space space space straight a over 3 space equals space straight b over 4 space equals space fraction numerator straight c over denominator negative 5 end fraction space equals space straight k space left parenthesis say right parenthesis$

∴ a = 3 k, b = 4 k, c = 5 k
Putting these values of a , b, c, in (1),
3 k (x – 2) + 4 k (y – 2) + (– 5 k) (z – 1) – 0
or 3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0
or 3 x – 6 + 4 y – 8 – 5 z + 5 = 0
or 3 x + 4 y – 5 z – 9 = 0
which is the required equation of plane.

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