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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.

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

The equation of line is $straight r with rightwards arrow on top space equals space 2 straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top space plus space straight lambda space left parenthesis 3 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$
or $straight x space straight i with hat on top space space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top space plus space straight lambda space left parenthesis 3 space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$
or $left parenthesis straight x minus 2 right parenthesis space straight i with hat on top space plus space left parenthesis straight y plus 1 right parenthesis space straight j with hat on top space plus space left parenthesis straight z minus 2 right parenthesis space straight k with hat on top space equals space straight lambda left parenthesis 3 straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis$

$therefore space space space space straight x minus 2 space space equals space 3 space straight lambda comma space space space straight y plus 1 space equals space 4 straight lambda comma space space space straight z minus 2 space equals space 2 space straight lambda therefore space space space space space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 4 end fraction space equals space fraction numerator straight z minus 2 over denominator 2 end fraction space equals space straight lambda$
Any point on this line is P(3λ + 2, 4λ –1, 2λ + 2)
It lies on the plane
$straight r with rightwards arrow on top. space left parenthesis straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top right parenthesis space equals space 5$
$straight i. straight e. space left parenthesis straight x straight i with hat on top space minus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top right parenthesis. space open parentheses straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 5 straight i. straight e. space space straight x minus straight y plus straight z minus 5 space equals space 0 therefore space space space space space 3 space straight lambda space plus space 2 space minus space 4 space straight lambda space plus space 1 space plus space 2 space straight lambda space plus space 2 minus 5 space equals 0 space space space space space space space space rightwards double arrow space space space space space straight lambda space equals space 0 therefore space space space point space straight P space is space left parenthesis 2 comma space minus 1 comma space 2 right parenthesis$
Required distance = distance between (–1,–5,–10) and (2, –1, 2)
$equals space square root of left parenthesis 2 plus 1 right parenthesis squared plus left parenthesis negative 1 plus 5 right parenthesis squared plus left parenthesis 2 plus 10 right parenthesis squared end root equals space square root of 9 plus 16 plus 44 end root space equals space square root of 169 space equals space 13 space units$

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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 $fraction numerator straight x plus 3 over denominator 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 6 end fraction space equals space straight z over 2$