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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.

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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.

The equations of line are

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

...(1)
Any point on this line is (3r + 2, 4r –11, 12 r + 2)
Let it lie on x – y + z = 5 ....(2)
∴ 3r + 2 – 4r + 1 + 12 r + 2 = 5 ⇒ 11r = 0 ⇒ r = 0
∴ point of intersection of line (1) and plane (2) is (2, – 1, 2)
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 space is space open parentheses negative 1 comma space minus 5 comma space minus 10 close parentheses.

Let d be required distance
∴ d = Distance between (2, – 1, 2) and (–1, – 5, – 10)

equals space square root of left parenthesis negative 1 minus 2 right parenthesis squared plus left parenthesis negative 5 plus 1 right parenthesis squared plus left parenthesis negative 10 minus 2 right parenthesis squared end root space equals space square root of 9 plus 16 plus 144 end root space equals space square root of 169 space equals space 13

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