JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If \[Q\left( 0,1,3 \right)\]is the image of the point P in the plane \[3xy+4z=2\]and R is the point (3, -1, -2), then the area (in sq. units) of \[\Delta PQR\]is: [JEE Main 10-4-2019 Morning]

    A) \[\frac{\sqrt{65}}{2}\]                       

    B) \[\frac{\sqrt{91}}{4}\]

    C) \[2\sqrt{13}\]              

    D) \[\frac{\sqrt{91}}{2}\]

    Correct Answer: D

    Solution :

    R lies on the plane.             \[DQ=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}=\sqrt{\frac{13}{2}}\]\[\Rightarrow PQ=\sqrt{26}\] Now,\[RQ=\sqrt{9+1}=\sqrt{10}\] \[\Rightarrow RD=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}\] Hence,\[ar(\Delta PQR)=\frac{1}{2}\times \sqrt{26}\times \sqrt{\frac{7}{2}}=\sqrt{\frac{91}{2}}.\]


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