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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.

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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.

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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.

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

The equation of given plane 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 space 3

...(1)
The equation of given plane 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 space 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

...(2)
Now line (2) passes through the point (1, 1, 0) with position vector

straight i with hat on top space plus space straight j with hat on top

and is parallel to the vector

2 space 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.

Now plane (1) passes through the point with position vector

straight i with hat on top space plus space straight j with hat on top

open parentheses straight i with hat on top space plus space straight j with hat on top close parentheses space. 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 space 3

i.e. if 1 + 2 -0 = 3
i.e. if 3 = 3, which is true
Now,

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

is a vector normal to the plane (1). It will be perpendicular to the line (2) if it is perpendicular to

2 space 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

i.e. if

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 space space open parentheses 2 space straight i with hat on top space plus space straight j with hat on top space plus space space 4 space straight k with hat on top close parentheses space equals space 0

i.e. if 2 + 2 – 4 = 0
i.e. if 0 = 0, which is true
∴ plane (1) contains the line (2 )

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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 $straight r with rightwards arrow on top. space open parentheses 3 straight i with hat on top space plus space straight j with hat on top space minus space straight k with hat on top close parentheses space plus space 2 space equals 0$