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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.

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

Let O be the origin and OA = a, OB = b, OC = c
∴ the equation of the plane passing through A, B and C is

straight x over straight a plus straight y over straight b plus straight z over straight c equals 1 space space space space space space space or space space space straight x over straight a plus straight y over straight b plus straight z over straight c minus 1 space equals space 0

From the given condition,

fraction numerator open vertical bar 0 plus 0 plus 0 minus 1 close vertical bar over denominator square root of begin display style 1 over straight a squared end style plus begin display style 1 over straight b squared end style plus begin display style 1 over straight c squared end style end root end fraction space equals space straight p

rightwards double arrow space space space space space square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root space equals 1 over straight p space space space space space space space rightwards double arrow space space space space 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 space space space space space space... left parenthesis 1 right parenthesis

The equation of plane through A (a, 0, 0) parallel to YZ-plane is x = a ... (2)
The equation of plane through B (0, b, 0) parallel to ZX-plane is y = b ... (3)
The equation of plane through C (0, 0, c) parallel to XY-plane is z = c ... (4)
Now to find the locus, we are to eliminate a, b, c.
Putting the values of a, b, c from (2), (3), (4) in (1), we get,
$1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals 1 over straight p squared$