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JEE Main & Advanced Mathematics Vector Algebra Question Bank Application of vectors in three dimensional geometry

  • question_answer
    The equation of the plane passing through the points \[(-1,-2,\,0),(2,\,3,\,5)\] and parallel to the line                  \[\mathbf{r}=-3\mathbf{j}+\mathbf{k}+\mathbf{\lambda }(2\mathbf{i}+5\mathbf{j}-\mathbf{k})\] is [J & K 2005]

    A)            \[\mathbf{r}.(-30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4\]

    B)            \[\mathbf{r}.(30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4\]

    C)            \[\mathbf{r}.(30\mathbf{i}+13\mathbf{j}-5\mathbf{k})=4\]

    D)            \[\mathbf{r}.(30\mathbf{i}-13\mathbf{j}-5\mathbf{k})=4\]

    Correct Answer: A

    Solution :

               Let the equation of plane is \[a(x+1)+b(y+2)+c(z-0)=0\]                     ?..(i)                    As it passes through (2, 3, 5)                    so, \[3a+5b+5c=0\]                                      ?..(ii)                    also, \[2a+5b-c=0\]                                      ?..(iii)                    \[\therefore \]\[\frac{a}{-5-25}=\frac{b}{10+3}=\frac{c}{15-10}\]                    \[\therefore \] \[\frac{a}{-30}=\frac{b}{13}=\frac{c}{5}\]                    Hence equation of plane is, \[-30x+13y+5z=4\]                    or \[\mathbf{r}.(-30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4\].


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