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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.

The equation of plane is
3x – y – z = 7 ...(1)
Direction ratios of the normal to the plane are 3, – 1, – 1.
Let M be the foot of perpendicular from P (2, 2, 7) to the plane.
Now MP is a straight line which passes through P (2, 2, 7) and has direction ratios as 3, – 1, – 1.

$therefore$ its equations are $fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y minus 2 over denominator negative 1 end fraction space equals space fraction numerator straight z minus 7 over denominator negative 1 end fraction equals space straight r space left parenthesis say right parenthesis$
Any point on line is M (3 r + 2, – r + 2, – r + 7)
∵ M lies on plane (1)
∵ 3 (3 r + 2) – (– r + 2) – (– r + 7) = 7
∴ 9r + 6 + r – 2 + r – 7 = 7
$rightwards double arrow space space space space 11 straight r space equals space 10 space space space space space space space space space space space space space rightwards double arrow space space space space straight r space equals space 10 over 11 therefore space space space space straight M space is space open parentheses 30 over 11 plus 2 comma space space space minus 10 over 11 plus 2 comma space space minus 10 over 11 plus 7 close parentheses space space space or space space space open parentheses 52 over 11 comma space 12 over 11 comma space 67 over 11 close parentheses therefore space space space space foot space of space perpendicular space is space open parentheses 52 over 11 comma space 12 over 11 comma space 67 over 11 close parentheses Length space of space perpendicular space space equals space open vertical bar fraction numerator 3 left parenthesis 2 right parenthesis minus 2 minus 7 minus 7 over denominator square root of 9 plus 1 plus 1 end root end fraction close vertical bar space equals space fraction numerator 10 over denominator square root of 11 end fraction$

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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.

105 Views

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