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Three Dimensional Geometry

Long Answer Type

261. Find the coordinates of the foot of perpendicular from the point (2, 2, 7) to the plane 3 x – y – z = 7. Also find the length of perpendicular.

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262. Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.

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263. Find the Cartesian as well as the vector equation of the planes passing through the intersection of the planes $straight r with rightwards arrow on top space. left parenthesis 2 space straight i with hat on top space plus space 6 space straight j with hat on top right parenthesis space plus space 12 space equals space 0 space space and space straight r with rightwards arrow on top. space space open parentheses 3 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses space equals space 0 comma$ which are at unit distance from the origin.

The equation of first plane is

straight r with rightwards arrow on top space. space open parentheses 2 straight i with hat on top space plus space 6 straight j with hat on top close parentheses space plus space 12 space equals space 0

open parentheses 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 close parentheses. space open parentheses 2 space straight i with hat on top space plus space 6 space straight j with hat on top close parentheses space plus space 12 space equals space 0

or space space space 2 straight x plus 6 straight y plus 12 space equals space 0 comma space space space space space space or space space space straight x plus 3 straight y plus 6 space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

The equation of second plane is

straight r with rightwards arrow on top. space left parenthesis 3 straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space space equals 0

open parentheses 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 close parentheses space. space open parentheses 3 space straight i with hat on top space minus space straight j with hat on top space plus space space 4 space straight k with hat on top close parentheses space equals space 0

3 straight x minus straight y plus 4 straight z space equals space 0

...(2)
The equation of any plane through the intersection of planes (1) and (2) is
(x + 3 y + 6) + k (3 x – y + 4 z) = 0
or (3 k + l).x + (–k + 3)y + 4 kz + 6 = 0 ...(3)
From given condition,
perpendicular distance of origin (0, 0, 0) from plane (3) = 1

therefore space space space fraction numerator 0 plus 0 plus 0 plus 6 over denominator square root of left parenthesis 3 space straight k space plus space 1 right parenthesis squared plus left parenthesis negative straight k plus 3 right parenthesis squared plus left parenthesis 4 space straight k right parenthesis squared end root end fraction space equals plus-or-minus 1 rightwards double arrow space space space space space left parenthesis 3 straight k plus 1 right parenthesis squared plus space left parenthesis negative straight k plus 3 right parenthesis squared plus 16 space straight k squared space equals space 36 rightwards double arrow space space 9 straight k squared plus 6 straight k plus 1 plus straight k squared space minus space 6 straight k space plus space 9 space space plus 16 straight k squared space equals space 36 rightwards double arrow space space 26 straight k squared space equals space 26 space space space space rightwards double arrow space space space space straight k squared space equals space 1 space space space rightwards double arrow space space space straight k space equals space plus-or-minus 1

Taking k = 1, from (3), we get,

4 straight x plus 2 straight y plus 4 straight z plus 6 space equals space 0 comma space space space space or space space space space 2 straight x plus straight y plus 2 straight z plus 3 space equals space 0 or space space 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 open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses plus space 3 space space equals space 0 or space space straight r with rightwards arrow on top space. space left parenthesis 2 straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis space plus space 3 space equals space 0

Taking k = – 1, from (3), we get,

negative 2 straight x plus 4 space straight y minus 4 straight z plus 6 space equals space 0 comma space space space space or space space space straight x minus 2 straight y plus 2 straight z minus 3 space equals space 0

or space space 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 open parentheses straight i with hat on top space minus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space minus space 3 space equals 0

or space space space space straight r with rightwards arrow on top. space open parentheses straight i with hat on top space minus space space 2 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses minus 3 space equals space 0

The required cartesian equations are
2x + y + 2z + 3 = 0, x – 2y + 2z — 3 = 0
and vector equations are

straight r with rightwards arrow on top. space left parenthesis 2 space straight i with hat on top space plus space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis space plus space 3 space equals space 0 space space space and space space straight r with rightwards arrow on top. space space open parentheses straight i with hat on top space minus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space minus space 3 equals 0.

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Short Answer Type

264. Prove that if a plane has the intercepts a, b, c and is at a distance p units from the origin, then

1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared space equals space 1 over straight p squared.

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Long Answer Type

265. A variable plane which remains at a constant distance 3p from the origin, cuts the co-ordinate axes at A, B, C. Show that the locus of the centroid of the triangle ABC is

1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals 1 over straight p squared.

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266. A variable plane is at a constant distance p from the origin and meets the axes in A, B and C respectively, then show that locus of the centroid of the triangle ABC is

1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals space 9 over straight p squared.

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Short Answer Type

267. If a variable plane at a constant distance p from the origin meets the coordinate axes in points A, B and C respectively. Through these points, planes are drawn parallel to the coordinate planes. Then show that the locus of the point of intersection is