Show that the line whose vector equation is  lies in the plane
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    Three Dimensional Geometry

    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.

    The equation of 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 line is
      straight r with rightwards arrow on top 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 space 4 space straight k with hat on top close parentheses                        ...(2)
    Now the line (2) will lie in plane (1)
    (i) if the point with position vector  straight i with hat on top space plus space straight j with hat on top lies in the plane
    i.e.   if  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) (1) + (1) (2) + (0) (– 1) = 3
    i.e. if 1 + 2 + 0 = 3
    i.e. if 3 = 3, which is true.
    and (ii) 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 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 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 (1) (2) + (2) (1) + (– 1) (4) = 0
    i.e. if 2 + 2 – 4 = 0, which is true
    ∴ line (2) lies in plane (1).

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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
    105 Views
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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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