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Three Dimensional Geometry

Short Answer Type

161.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3 y – z = 5

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162.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0

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Long Answer Type

163.

In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin.
2x + 3y + 4z – 12 = 0

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164. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
3 y + 4 z – 6 = 0

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Short Answer Type

165. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1

The equation of plane is
x + y + z = 1 ...(1)
Dividing both sides by

square root of left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared end root space equals space square root of 3 comma space space space we space get comma

fraction numerator 1 over denominator square root of 3 end fraction straight x plus fraction numerator 1 over denominator square root of 3 end fraction straight y plus fraction numerator 1 over denominator square root of 3 end fraction straight z space equals space fraction numerator 1 over denominator square root of 3 end fraction

∴ direction ratios of the normal OP to the plane are

fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma space fraction numerator 1 over denominator square root of 3 end fraction comma

where O is origin and P (x₁, y₁, z₁) is foot of perpendicular.
Direction ratios of OP are x₁ – 0, y₁ – 0, z₁– 0 i.e. x₁, y₁, z₁.
Since direction cosines and direction ratios of a line are proportional.

therefore space space space space space space space space fraction numerator straight x subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals space fraction numerator straight y subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals fraction numerator straight z subscript 1 over denominator begin display style fraction numerator 1 over denominator square root of 3 end fraction end style end fraction space equals space straight k comma space space say therefore space space space space space space space space straight x subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k. space space space space space straight y subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k comma space space space space straight z subscript 1 space equals space fraction numerator 1 over denominator square root of 3 end fraction straight k therefore space space space space straight P space is space open parentheses fraction numerator 1 over denominator square root of 3 end fraction straight k comma space fraction numerator 1 over denominator square root of 3 end fraction straight k comma space fraction numerator 1 over denominator square root of 3 end fraction straight k close parentheses

Since P lies on plane (1)

therefore space space space space fraction numerator straight k over denominator square root of 3 end fraction plus fraction numerator straight k over denominator square root of 3 end fraction plus fraction numerator straight k over denominator square root of 3 end fraction space equals space 1 space space space or space space straight k plus straight k plus straight k space equals space square root of 3 space space space or space space 3 space straight k space equals space square root of 3 space space rightwards double arrow space space straight k space equals space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space straight P space is space open parentheses 1 third comma space 1 third comma space 1 third close parentheses comma space which space is space foot space of space perpendicular. space

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166. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
5y + 8 = 0

97 Views

167. Find the direction cosines of the perpendicular from the origin to the plane

straight r with rightwards arrow on top. space open parentheses 6 space straight i with hat on top space minus space 3 space straight j with hat on top space minus space 2 space straight k with hat on top close parentheses space plus space 1 space equals 0

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168. Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector

2 space straight i with hat on top space plus space 6 space straight j with hat on top space minus space 3 space straight k with hat on top

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169. Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector

3 straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top.

86 Views

Long Answer Type

170. Find the vector equation of the plane which is at a distance of

fraction numerator 6 over denominator square root of 29 end fraction

from the origin and its normal vector from the origin is

2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space space 4 space straight k with hat on top.

Also find its cartesian form.

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Previous Year Papers

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Test Yourself

Short Answer Type

Long Answer Type

Short Answer Type

165. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.x + y + z = 1

Long Answer Type

165. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
x + y + z = 1