Show that the line  lies in the plane  from Mathematics Thre

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 Multiple Choice QuestionsLong 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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 Multiple Choice QuestionsShort 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
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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


The equation of given plane is
    straight r with rightwards arrow on top. space open parentheses 3 space 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 equals space minus 2
or  straight r with rightwards arrow on top. space open parentheses negative 3 space 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 minus 2                                             ...(1)
The equation of given line is
straight r with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top space plus space straight lambda open parentheses straight i with hat on top space minus space straight j with hat on top space plus space space 2 space straight k with hat on top close parentheses                          ...(2)
Now line  (2) passes through the point (2, -3, 5), With position vector
2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top and is parallel to the vector 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.
Now plane (1) passes through the point with position vector 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top
if   open parentheses 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 5 space straight k with hat on top close parentheses. space space open parentheses negative 3 space 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 2
i.e.   if (2) (-3) + (-3) (-1) + (5) (1) = 2
i.e.  if 2 = 2, which is true
Now, negative 3 straight i with hat on top space minus space straight j with hat on top space plus 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 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       
i.e. if (–3) (1) + (– 1) (– 1) + (1) (2) = 0
i.e. if –3 + 1+ 2 = 0
i.e. if 0 = 0, which is true
∴  line (2) lies in plane (1).
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296. Show that the line whose vector equation is straight r with rightwards arrow on top space space equals space open parentheses straight i with hat on top space plus space straight j with hat on top close parentheses space plus space straight lambda space open parentheses 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 close parentheses lies in the plane straight pi 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 space 3.
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297. Find the equations of the line passing through (1, – 2, 3) and parallel to the planes x – y + 2 z = 5 and 3 x + 2y – z = 6. 
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 Multiple Choice QuestionsLong Answer Type

298. Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes straight r with rightwards arrow on top. space open parentheses 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 close parentheses space equals space 5 space space and space straight r with rightwards arrow on top. space space open parentheses 3 space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top close parentheses space equals space 6.
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299. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the lines:
fraction numerator straight x minus 8 over denominator 3 end fraction space equals space fraction numerator straight y plus 19 over denominator negative 16 end fraction space equals space fraction numerator straight z minus 10 over denominator 7 end fraction space and space fraction numerator straight x minus 15 over denominator 3 end fraction space equals space fraction numerator straight y minus 29 over denominator 8 end fraction space equals space fraction numerator straight z minus 5 over denominator negative 5 end fraction
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300. Find the equation of the plane containing the line
fraction numerator straight x plus 2 over denominator 2 end fraction space equals space fraction numerator straight y plus 3 over denominator 3 end fraction space equals fraction numerator straight z minus 4 over denominator negative 2 end fraction
and the point (0, 6, 0).
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