JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line

  • question_answer
    The point of intersection of the lines \[\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1},\] \[\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\] is [MP PET 2004]

    A)                                                               \[21,\,\frac{5}{3},\frac{10}{3}\]

    B)                                                               \[(\,2,\,10,\,4)\]

    C)                                                               \[(-3,\,3,\,6)\]

    D)                                                               \[(5,\,7,\,-2)\]

    Correct Answer: A

    Solution :

                    Given lines are,                                                 \[\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}={{r}_{1}}\], (say)                                 and        \[\frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}={{r}_{2}}\], (say)                                 \[\therefore \]\[x=3{{r}_{1}}+5=-36{{r}_{2}}-3\], \[y=-{{r}_{1}}+7=3+2{{r}_{2}}\] and \[z={{r}_{1}}-2=4{{r}_{2}}+6\]                                 On solving, we get \[x=21,\,y=\frac{5}{3},\,z=\frac{10}{3}\].                                                                       Trick: Check through options.


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