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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    The vector equation of the plane through the line of intersection of the planes \[x+y+z=1\]and \[2x+3y+4z=5\]which is perpendicular to the plane \[xy+z=0\] is :             [JEE Main 8-4-2019 Afternoon]

    A) \[\vec{r}\times (\hat{i}+\hat{k})+2=0\]      

    B) \[\vec{r}.(\hat{i}-\hat{k})-2=0\]

    C) \[\vec{r}.(\hat{i}-\hat{k})+2=0\]

    D)   \[\vec{r}\times (\hat{i}-\hat{k})+2=0\]

    Correct Answer: C

    Solution :

    Let the plane be \[\left( x+y+z1 \right)+\lambda \left( 2x+3y+4z5 \right)=0\] \[\Rightarrow \]\[(2\lambda +1)x+(3\lambda +1)y+(4\lambda +1)z-(5\lambda +1)=0\] \[\bot \]to the plane \[xy+z=0\] \[\Rightarrow \]\[\lambda =-\frac{1}{3}\] \[\Rightarrow \] the required plane is \[xz+2=0\]


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