Show that the lines:            and        are copla
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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

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    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. 

    The equations of given lines are
                              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
    These lines are coplanar
        if            open vertical bar table row cell straight b minus straight c minus straight a plus straight d end cell cell straight b minus straight a end cell cell straight b plus straight c minus straight a minus straight d end cell row cell straight a minus straight delta end cell straight alpha cell straight alpha plus straight delta end cell row cell straight beta minus straight gamma end cell straight beta cell straight beta plus straight gamma end cell end table close vertical bar space equals space 0


    i.e. if        open vertical bar table row cell 2 straight b minus 2 straight a end cell cell straight b minus straight a end cell cell straight b plus straight c minus straight a minus straight d end cell row cell 2 straight alpha end cell straight alpha cell straight alpha plus straight delta end cell row cell 2 straight beta end cell straight beta cell straight beta plus straight gamma end cell end table close vertical bar space equals space 0 comma space space by space straight C subscript 1 plus straight C subscript 3

    i.e. if   open vertical bar table row cell straight b minus straight a end cell cell straight b minus straight a end cell cell straight b plus straight c minus straight a minus straight d end cell row straight alpha straight alpha cell straight alpha plus straight delta end cell row straight beta straight beta cell straight beta plus straight gamma end cell end table close vertical bar space equals 0             

    i.e. if    2(0) = 0    [Since C1, C2 are identical]
    i.e. if    0 = 0, which is true.
    Hence the result.

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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. 
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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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