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Three Dimensional Geometry

Long Answer Type

291. Show that the lines $fraction numerator straight x minus 1 over denominator 3 end fraction space equals space fraction numerator straight y plus 1 over denominator 2 end fraction space equals fraction numerator straight z minus 1 over denominator 5 end fraction space space and space fraction numerator straight x minus 2 over denominator 4 end fraction space equals space fraction numerator straight y minus 1 over denominator 3 end fraction space equals space fraction numerator straight z plus 1 over denominator negative 2 end fraction$ do not intersect each other.

The equations of lines are

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

...(1)

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

...(2)
Any point on the line (1) is (3r + 1, 2r – 1, 5r + 1)
It lies on line (2)
if

fraction numerator 3 straight r plus 1 minus 2 over denominator 4 end fraction space equals space fraction numerator 2 straight r minus 1 minus 1 over denominator 3 end fraction space equals space fraction numerator 5 straight r plus 1 plus 1 over denominator negative 2 end fraction

i.e., if

fraction numerator 3 straight r minus 1 over denominator 4 end fraction space equals space fraction numerator 2 straight r minus 2 over denominator 3 end fraction space equals space fraction numerator 5 straight r plus 2 over denominator negative 2 end fraction

...(3)

From the first and second numbers,

fraction numerator 3 straight r minus 1 over denominator 4 end fraction space equals space fraction numerator 2 straight r minus 2 over denominator 3 end fraction space space space space space space space or space space space space 9 straight r minus 3 space equals space 8 or space space space straight r space equals space minus 5

Substituting this value of r in (3), we get,

fraction numerator negative 15 minus 1 over denominator 4 end fraction space equals space fraction numerator negative 10 minus 2 over denominator 3 end fraction space equals fraction numerator negative 25 plus 2 over denominator negative 2 end fraction

negative 4 space equals space minus 4 space space equals 23 over 2 comma

which is not true
∴ the given lines do not intersect.

91 Views

Short Answer Type

292. Show that the lines:

fraction numerator straight x minus straight a plus straight d over denominator straight alpha minus straight delta end fraction space equals space fraction numerator straight y minus straight a over denominator straight alpha end fraction space equals space fraction numerator straight z minus straight a minus straight d over denominator straight a plus straight delta end fraction

and

fraction numerator straight x minus straight b plus straight c over denominator straight beta minus straight gamma end fraction space equals space fraction numerator straight y minus straight b over denominator straight beta end fraction space equals space fraction numerator straight z minus straight b minus straight c over denominator straight beta plus straight gamma end fraction

are coplanar.

73 Views

293. If 4x + 4y – kz = 0 is the equation of the plane through the origin that contains a line

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

then find the value of k.

100 Views

294. Show that the plane whose vector equation is

straight r with rightwards arrow on top. space open parentheses straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top close parentheses space equals 3

contains the line whose vector equation is

straight r with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight lambda open parentheses 2 straight i with hat on top space plus space straight j with hat on top space plus space 4 space straight k with hat on top close parentheses

102 Views

295.

Show that the line $straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus 3 space space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight lambda space open parentheses straight i with hat on top space minus space straight j with hat on top space plus 2 space straight k with hat on top close parentheses$ lies in the plane $straight r with rightwards arrow on top. space open parentheses 3 straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top close parentheses space plus space 2 space equals 0$