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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines \[\vec{r}=\left( \hat{i}+\hat{j} \right)+\lambda \left( i+2j-k \right)\]and \[\vec{r}=\left( \hat{i}+\hat{j} \right)+\mu \left( -\hat{i}+\hat{j}-2\hat{k} \right)\]is : [JEE Main 12-4-2019 Afternoon]

    A) \[\sqrt{3}\]      

    B) \[\frac{1}{\sqrt{3}}\]

    C) \[\frac{1}{3}\]                        

    D) \[3\]

    Correct Answer: A

    Solution :

                perpendicular vector to the plane \[\vec{n}\left| \begin{matrix}    i & j & k  \\    1 & 2 & -1  \\    -1 & 1 & -2  \\ \end{matrix} \right|=-3\hat{i}+3\hat{j}+3\hat{k}\] Eq. of plane \[-3\left( x1 \right)+3\left( y1 \right)+3z=0\] \[\Rightarrow xyz=0\] \[{{d}_{(2,1,4)}}=\frac{\left| 2-1-4 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}}=\sqrt{3}\]           


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