Find the equations of the line passing through (1, – 2, 3) and

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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

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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. 


Let a, b, c the direction ratios of the line passing through the point (1, - 2, 3).
∴    equations of line are
           fraction numerator straight x minus 1 over denominator straight a end fraction equals fraction numerator straight y plus 2 over denominator straight b end fraction equals fraction numerator straight z minus 3 over denominator straight c end fraction                                              ...(1)
Since line (1) parallel to the plane x - y + 2z = 5 whose direction ratios of the normal are 1, -1, 2
∴  a(1) +b(– 1) + c(2) = 0
or    a – b +2c=0    ... (2)
Again line (1) is parallel to the plane 3x + 2y – z = 6
∴ 3a + 2b – c = 0    ...(3)
From (2) and (3), we get
                   fraction numerator straight a over denominator 1 minus 4 end fraction space equals space fraction numerator straight b over denominator 6 plus 1 end fraction space equals space fraction numerator straight c over denominator 2 plus 3 end fraction
therefore space space space space fraction numerator straight a over denominator negative 3 end fraction space equals space straight b over 7 space equals space straight c over 5
∴ from (1), the equations of line are
fraction numerator straight x minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight y plus 2 over denominator 7 end fraction space equals space fraction numerator straight z minus 3 over denominator 5 end fraction

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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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