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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

The equation of plane is
5y + 8 = 0 or 5y = – 8
or 0x – 5 y + 0z = 8
Dividing both sides by $square root of left parenthesis 0 right parenthesis squared plus left parenthesis 5 right parenthesis squared plus left parenthesis 0 right parenthesis squared end root space equals space 5 comma space space space we space get comma$
0x - y + 0 z = $8 over 5$
It is of the form lx + my + nz = p, where l = 0, m = -1, n = 0, $straight p space equals space 8 over 5$
∴ direction cosines of the normal to the plane are 0, – 1, 0 and distance from origin = $8 over 5.$

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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

78 Views

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

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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

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

Long Answer Type

Short Answer Type

Long Answer Type

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