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JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    If the line \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\] and \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\] intersect, then k is equal to :   AIEEE  Solved  Paper-2012

    A) \[-1\]                                       

    B) \[\frac{2}{9}\]

    C) \[\frac{9}{2}\]                                      

    D) 0

    Correct Answer: C

    Solution :

                 \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\]           \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\] \[\vec{a}\,(1,-1,1);\]                          \[\vec{r}\,=\vec{a}+\lambda \vec{b}\] \[\vec{b}\,(2,3,4)\] \[\vec{c}\,(3,k,0);\]            \[\vec{r}=\vec{a}+\mu \vec{d}\] \[\vec{d}(1,2,1)\] These lines will intersect if lines are coplaner                    \[\vec{a}-\vec{c},\,\,\hat{b}\And \vec{d}\] are coplaner \[\therefore \]     \[[\vec{a}-\vec{c},\vec{b},\vec{d}]=0\]                    \[\left| \begin{matrix}    2 & k+1 & -1  \\    2 & 3 & 4  \\    1 & 2 & 1  \\ \end{matrix} \right|=0\] \[\Rightarrow 2(-5)-(k+1)(-2)-1(1)=0\] \[\Rightarrow \,2(k+1)=11\] \[\Rightarrow k=\frac{9}{2}\]


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