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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    Locus of all the points which are at a distance of 3 units from the line \[\overset{\scriptscriptstyle\smile}{r}=\lambda (\hat{i}+\hat{j}+\hat{k})\] is given by \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx=k.\]The value of k is

    A) \[18\]                    

    B)        \[27\]

    C) \[27/2\]               

    D)        \[9/2\]

    Correct Answer: C

    Solution :

      [c] Given line is \[\vec{r}=\lambda (\hat{i}+\hat{j}+\hat{k})\] Any point on the line is \[Q(\lambda ,\lambda ,\lambda )\] Let point \[P(x,y,z)\] be at a distance of 3 units from point Q. \[\therefore \,\,\,\,\,P{{Q}^{2}}=9\] \[\Rightarrow \,\,\,{{(x-\lambda )}^{2}}+{{(y-\lambda )}^{2}}+{{(2-\lambda )}^{2}}=9\]   ?..(1) Also, PQ is perpendicular to the given line. \[\therefore \,\,\overrightarrow{QP}.\left( \hat{i}+\hat{j}+\hat{k} \right)=0\] \[\Rightarrow \,\,\,\,x-\lambda +y-\lambda +z-\lambda =0\] \[\Rightarrow \,\,\,\,\lambda =\frac{x+y+z}{3}\] Putting this value of \[\lambda ,\] in equation (1), we get locus of point Pas \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx=27/2\]


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