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Three Dimensional Geometry

Short Answer Type

51. Find a vector equation of the line through the points A (3, 4, – 7) and B (1, – 1, 6).

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52. Find the vector equation for the line passing through the points (– 1, 0, 2) and (3, 4, 6).

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53. Find the vector and cartesian equations of the line that passes through the origin and (5,- 2, 3).

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54. Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

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55. Find the vector equation of line joining the points whose vectors are

2 space straight i with hat on top minus straight j with hat on top plus straight k with hat on top space space and space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top.

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56. If the points (–1, 3, 2), (– 4, 2, –2) and (5, 5, λ) are collinear, then find the value of λ.

The equation of line through (–1,3, 2) and (– 4, 2, –2) is

fraction numerator straight x minus left parenthesis negative 1 right parenthesis over denominator negative 4 plus 1 end fraction space equals space fraction numerator straight y minus 3 over denominator 2 minus 3 end fraction space equals space fraction numerator straight z minus 2 over denominator negative 2 minus 2 end fraction

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

...(1)

Since points (–1, 3, 2), (– 4, 2, –2) and (5, 5, λ) are collinear
∴ point (5, 5, λ) lies on (1)

therefore space space space space space space space space space space fraction numerator 5 plus 1 over denominator negative 3 end fraction space equals space fraction numerator 5 minus 3 over denominator negative 1 end fraction space equals space fraction numerator straight lambda minus 2 over denominator negative 4 end fraction therefore space space space space space space space space space space space minus 2 space equals space space minus 2 space equals space fraction numerator straight lambda minus 2 over denominator negative 4 end fraction therefore space space space space space space space space straight lambda minus 2 space equals space 8 space space space space space space space rightwards double arrow space space space space straight lambda space equals space 10

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Long Answer Type

57. Show that the points whose position vectors are given by

2 space straight i with hat on top space plus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space minus space 4 space straight i with hat on top space plus space 3 space straight j with hat on top space minus space straight k with hat on top comma space space 5 space straight i with hat on top space plus space 5 space straight k with hat on top

are collinear.

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Short Answer Type

58.

Show that the point whose position vectors are given by $negative 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top comma space space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top space space and space space 7 straight i with hat on top space minus space straight k with hat on top$ are collinear.

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Long Answer Type

59. Find the coordinates of the point where the line through A(3, 4, 1) and B (5, 1, 6) crosses the x y-plane.

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60. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

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