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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

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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

The equation of plane is
5 y + 8 = 0 or 5 y = – 8 or – 5 y = 8
Dividing both sides by $square root of left parenthesis 0 right parenthesis squared plus left parenthesis negative 5 right parenthesis squared plus left parenthesis 0 right parenthesis squared end root space equals space 5 comma space we space get comma space space straight y space equals space 8 over 5$ ...(1)
∴ direction ratios of the normal OP to the plane are 0, – 1, 0 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.
$therefore space space space space space space straight x subscript 1 over 0 space equals space fraction numerator straight y subscript 1 over denominator negative 1 end fraction space equals space straight z subscript 1 over 0 space equals space straight k comma space space space say.$

$therefore space space space space space space space straight x subscript 1 space equals space 0 comma space space space space straight y subscript 1 space equals space minus straight k comma space space space straight z subscript 1 space equals space 0$

$therefore space space space space straight P space is space left parenthesis 0 comma space space minus straight k comma space space 0 right parenthesis$

Since P lies on plane (1)
$therefore space space space space space space straight k space equals space 8 over 5 therefore space space space space space straight P space is space open parentheses 0 comma space space minus 8 over 5 comma space 0 close parentheses.$