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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).

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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.

The equation of any plane through the intersection of planes
x + 2y + z – 3 = 0 and 2x – y – z – 5 = 0 is
(x + 2y + z – 3) + k (2x – y – z – 5) = 0 ...(1)
∴ it passes through (2, 1, 3)
∴ (2 + 2 + 3 – 3) + k (4 – 1– 3 – 5) = 0
$therefore space space space 4 space minus space 5 space straight k space equals space 0 space space space space space space space space space space space space rightwards double arrow space space space space straight k space equals space 4 over 5$
Putting $straight k space equals 4 over 5 space in space left parenthesis 1 right parenthesis comma space we space get comma$
$left parenthesis straight x plus 2 straight y plus straight z minus 3 right parenthesis plus space 4 over 5 left parenthesis 2 straight x minus straight y minus straight z minus 3 right parenthesis space equals space 0$
or 5 (x + 2 y + z – 3) + 4 (2 x – y – z – 3) = 0
or 5 x + 10 y + 5 z – 15 + 8 x – 4 y – 4 z – 12 = 0
or 13 x + 6 y + z – 27 = 0
which is equation of plane.
Direction ratios of normal to the plane are 13, 6, 1.

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