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Three Dimensional Geometry

Long Answer Type

231.

Find the image of the point (1, 3, 4) in the plane x – y + z = 5.

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Short Answer Type

232. A variable plane passes through a fixed point (a, b, c) and meets the co– ordinate axes in A, B. C. Show that the locus of the point common to the planes through A. B, C parallel to the co-ordinate planes is

straight a over straight x plus straight b over straight y plus straight c over straight z equals 1.

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Long Answer Type

233. Establish by vector method the equation of the plane making intercepts a. b, c on the co-ordinate axes in the form of

straight x over straight a plus straight y over straight b plus straight z over straight c space equals space 1

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Multiple Choice Questions

234. The planes: 2x – y + 4 z = 5 and 5.x – 2.5 y + 10 z = 6 are

Perpendicular
Parallel
intersect y-axis
intersect y-axis

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Short Answer Type

235. Find the distance of the point (3, 4, 5) from the plane $straight r with rightwards arrow on top. space left parenthesis 2 space straight i with hat on top space minus space 5 space straight j with hat on top plus space 3 space straight k with hat on top right parenthesis space equals 13$

The equation of plane is

straight r with rightwards arrow on top. space left parenthesis 2 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 13

or space left parenthesis straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space space left parenthesis 2 space straight i with hat on top space minus space 5 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 13 or space space space 2 straight x minus 5 straight y plus 3 straight z space equals space 13 or space space space 2 straight x minus 5 straight y space plus space 3 straight z minus 13 space equals space 0

Let p be the length of perpendicular from point (3, 4, 5) to the plane.

therefore space space space straight p space space equals space fraction numerator open vertical bar 2 space left parenthesis 3 right parenthesis space minus space 5 space left parenthesis 4 right parenthesis space plus space 3 space left parenthesis 5 right parenthesis space space minus space 13 close vertical bar over denominator square root of 4 plus 25 plus 9 end root end fraction space equals space fraction numerator open vertical bar 6 minus 20 plus 15 minus 13 close vertical bar over denominator square root of 38 end fraction space equals space fraction numerator 12 over denominator square root of 38 end fraction

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236. Find the distance from P (2, 1, – 1 ) to the plane x – 2 y + 4 z = 9.

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237. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(0, 0, 0) 3 x - 4 y+12 z = 3

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238. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(3, -2, 1) 2x - y + 2z + 3 = 0

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239. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(2, 3, -5) x + 2y - 2z = 9

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240. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(–6, 0, 0) 2x - 3y + 6z - 2 = 0

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