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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    The perpendicular bisector of the line segment with endpoints (2, 3, 2) and \[(-4,\,\,1,\,\,4)\] passes through the point \[(-3,\,\,6,\,\,1)\]) and has equation of, the form \[\frac{x+3}{a}=\frac{y-6}{b}=\frac{z-1}{c}\] where a, b and c are relatively prime integers with \[a>0\]. The value of abc \[-(a+b+c)\] is equal to

    A)  \[-1\]                           

    B)  \[-2\]

    C)  \[-3\]                           

    D)  \[-4\]

    Correct Answer: B

    Solution :

    The midpoint of the segment is \[\left( \frac{2-4}{2},\,\frac{3+1}{2},\,\frac{2+4}{2} \right)\]  = (-1, 2, 3) and the vector between this point and (-3, 6, 1) is <-1-(-3), 2 - 6, 3-1 > and<2,-4,2>. \[\therefore \]the line has parametric equations x=-3+2t, y = 6 - 4t and  z =1 + 2t. Solving each of these equations for t and setting the expressions equal makes the equation. \[\frac{x+3}{a}=\frac{y-6}{b}=\frac{z-1}{c}\] \[\Rightarrow \,\frac{x+3}{1}=\frac{y-6}{-2}\,=\frac{z-6}{1}\] So 1(-2)-1-(1-2+1)= - 2


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