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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.

Let A (– 1, 2, 3), B(3, –5, 6) be given points and C be mid-point of AB.

therefore space space space space space straight C space is space open parentheses fraction numerator negative 1 plus 3 over denominator 2 end fraction comma space fraction numerator 2 minus 5 over denominator 2 end fraction comma space fraction numerator 3 plus 6 over denominator 2 end fraction close parentheses

i.e.

open parentheses 1 comma space minus 3 over 2 comma space 9 over 2 close parentheses

Since the plane passes through the point

open parentheses 1 comma space minus space 3 over 2 comma space 9 over 2 close parentheses

∴ equation of plane is

straight A left parenthesis straight x minus 1 right parenthesis space plus space straight B space open parentheses straight y plus 3 over 2 close parentheses space plus space straight C open parentheses straight z minus 9 over 2 close parentheses space equals space 0

...(1)
Direction ratios of AB are 3 + 1, –5 – 2, 6 – 3 i.e. 4, –7, 3.
∴ the line with direction ratios 4, –7, 3 is normal to the plane (1)

therefore space space space space 4 left parenthesis straight x minus 1 right parenthesis space minus space 7 space open parentheses straight y plus 3 over 2 close parentheses space plus space space 3 space open parentheses straight z minus 9 over 2 close parentheses space equals 0 therefore space space 4 straight x minus 4 minus 7 straight y minus 21 over 2 plus 3 straight z minus 27 over 2 space equals space 0 or space space space 4 straight x minus 7 straight y space plus space 3 straight z space minus space 28 space equals space 0

which is required equation of plane.

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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).

87 Views

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.

99 Views

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