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Three Dimensional Geometry

Short Answer Type

271. Find the distance of a point P (5, 3, 4) from the point where the line

fraction numerator straight x minus 3 over denominator 1 end fraction space equals fraction numerator straight y minus 4 over denominator 2 end fraction space equals fraction numerator straight z minus 5 over denominator 2 end fraction

meets the plane x + y + z = 2.

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272. Find the distance of a point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line

straight x over 2 space equals space straight y over 3 space equals space fraction numerator straight z over denominator negative 6 end fraction

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273. Find the distance of the point (- 2, 3, - 4) from the line $fraction numerator straight x plus 2 over denominator 3 end fraction space equals space fraction numerator 2 straight y plus 3 over denominator 4 end fraction space equals space fraction numerator 3 straight z plus 4 over denominator 5 end fraction$ measured parallel to the plane 4 x + 12 y - 3z + 1 = 0.

The equations of given line are

fraction numerator straight x plus 2 over denominator 3 end fraction space equals space fraction numerator 2 straight y plus 3 over denominator 4 end fraction space equals space fraction numerator 3 straight z plus 4 over denominator 5 end fraction

fraction numerator straight x plus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus begin display style 3 over 2 end style over denominator 2 end fraction space equals space fraction numerator straight z plus begin display style 4 over 3 end style over denominator begin display style 5 over 3 end style end fraction space equals space straight r space left parenthesis say right parenthesis

Any point Q on this line is

open parentheses 3 straight r minus 2 comma space space 2 straight r minus 3 over 2 comma space 5 over 3 straight r space minus space 4 over 3 close parentheses

Let given point be P(– 2, 3, – 4). Direction ratios of PQ are

3 straight r minus 2 plus 2 comma space space space 2 straight r minus 3 over 2 minus 3 comma space space 5 over 3 straight r minus 4 over 3 plus 4 space space straight i. straight e. space 3 straight r comma space space 2 straight r minus 9 over 2 comma space space fraction numerator 5 straight r over denominator 3 end fraction plus 8 over 3

Since PQ is parallel to the plane

4 straight x plus 12 straight y minus 3 straight z plus 1 space equals space 0

therefore space 4 space left parenthesis 3 straight r right parenthesis space plus space 12 space open parentheses 2 straight r minus space 9 over 2 close parentheses space plus space left parenthesis negative 3 right parenthesis space open parentheses fraction numerator 5 straight r over denominator 3 end fraction plus 8 over 3 close parentheses space equals space 0 therefore space space space 12 straight r space plus space 24 straight r space minus space 54 space minus space 5 straight r space minus space 8 space equals space 0 space therefore space space space space space space space space space 31 space straight r space equals space 62 space space space space rightwards double arrow space space space space straight r space equals space 2 therefore space space space space space space space space space space space straight Q space is space open parentheses 6 minus 2 comma space space 4 minus 3 over 2 comma space space 10 over 3 minus 4 over 3 close parentheses space straight i. straight e. comma space space space space space open parentheses 4 comma space 5 over 2 comma space 2 close parentheses

∴ required length = PQ

equals space square root of left parenthesis 4 plus 2 right parenthesis squared plus open parentheses 5 over 2 minus 3 close parentheses squared plus left parenthesis 2 plus 4 right parenthesis squared end root equals space square root of 36 plus 1 fourth plus 36 end root space equals space square root of fraction numerator 144 plus 1 plus 144 over denominator 4 end fraction end root space equals space square root of 289 over 4 end root space equals space 17 over 2 space units. space

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Long Answer Type

274. Find the distance of the point (–1, –5, –10) from the point of intersection of the line

straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top space plus space straight lambda space open parentheses 3 straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top close parentheses space and space the space plane space straight r with rightwards arrow on top. space open parentheses straight i with hat on top minus straight j with hat on top space plus space space straight k with hat on top close parentheses space equals space 5.

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Short Answer Type

275. Find the distance between the point with position vector

negative straight i with hat on top space minus space 5 space straight j with hat on top space minus space 10 space straight k with hat on top

and the point of intersection of the line

fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 2 over denominator 12 end fraction

with the plane x – y + z = 5.

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276.

Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line $fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 6 end fraction space equals space straight z over 2$