Find the vector equation of a plane which is at a distance of 7

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 Multiple Choice QuestionsShort 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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 Multiple Choice QuestionsLong 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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 Multiple Choice QuestionsShort 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.


Here   p = 7
and    straight n with rightwards arrow on top space equals space 3 space 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
therefore     straight n with hat on top space equals space fraction numerator straight n with rightwards arrow on top over denominator open vertical bar straight n with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator 3 space 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 over denominator square root of 9 plus 25 plus 36 end root end fraction space equals space fraction numerator 1 over denominator square root of 70 end fraction left parenthesis 3 space 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 right parenthesis
The vector equation of plane is
                             straight r with rightwards arrow on top. space straight n with hat on top space equals space straight p
or               straight r with rightwards arrow on top. space open parentheses fraction numerator 3 space 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 over denominator space square root of 70 end fraction close parentheses space equals space 7.

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 Multiple Choice QuestionsLong 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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