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Three Dimensional Geometry

Short Answer Type

191. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

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192. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

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193. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

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194. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them:
4x + 8y + z – 8 = 0 and y + z – 4 = 0

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195. Find the equation of the plane which bisects the line joining the points (–1, 2, 3) and (3, – 5 6) at right angles.

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196. Find the equation of the plane through the intersection of the planes x + y + z = 9 and 2 x + 3 y + 4 z + 5 = 0 and passing through the point (1, 1, 1).

The equation of any plane through the intersection of planes
x + y + z = 9 i.e., x + y + z – 9 = 0 and 2 x + 3 y + 4 z + 5 = 0 is
(x + y + z – 9) + k (2 x + 3 y + 4 z + 5) = 0    ...(1)
∴ it passes through the point (1, 1, 1)
∴ (1 + 1 + 1 – 9) + k (2 + 3 + 4 + 5) = 0
$rightwards double arrow space space space minus 6 space plus space 14 space straight k space equals space 0 space space space space space space rightwards double arrow space space space space straight k space equals space 6 over 14 space equals space 3 over 7$
Putting this value of k in (1), we get,
   $open parentheses straight x plus straight y plus straight z minus 9 close parentheses plus 3 over 7 left parenthesis 2 straight x plus 3 straight y plus 4 straight z plus 5 right parenthesis space equals 0$
or    $7 space left parenthesis straight x plus straight y plus straight z minus 9 right parenthesis space plus space 3 space left parenthesis 2 straight x plus 3 straight y plus 4 straight z plus 5 right parenthesis space equals space 0$
or $7 straight x plus 7 straight y plus 7 straight z minus 63 plus 6 straight x plus 9 straight y plus 12 straight z plus 15 space equals space 0$
or    $13 straight x plus 16 straight y plus 19 straight z space minus space 48 space equals 0$
which is required equation of plane.

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Long Answer Type

197. Find the equation of plane passing through origin and intersection of planes 2x – 3y + z = 9, x – y + z = 4

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Short Answer Type

198. Find the equation of the plane through the intersection of the planes 3x – y + 2 z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

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199. Find the equation of the plane passing through the point (– 1, – 1, 2) and perpendicular to each of the following planes:
2x + 3y – 3 = 2 and 5x – 4y + z = 6.

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Long Answer Type

200. Find the direction ratios of the normal to the plane passing through the point (2,1, 3) and the line of intersection of the planes x + 2 y + z = 3 and 2 x – y – z = 5.

334 Views