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

  • question_answer
    The position vector of a point at a distance of \[3\sqrt{11}\] units from \[\mathbf{i}-\mathbf{j}+2\mathbf{k}\] on a line passing through the points \[\mathbf{i}-\mathbf{j}+2\mathbf{k}\] and \[3\mathbf{i}+\mathbf{j}+\mathbf{k}\] is                                

    A)            \[10\mathbf{i}+2\mathbf{j}-5\mathbf{k}\]

    B)            \[-8\mathbf{i}-4\mathbf{j}-\mathbf{k}\]

    C)            \[8\mathbf{i}+4\mathbf{j}+\mathbf{k}\]

    D)            \[-10\mathbf{i}-2\mathbf{j}-5\mathbf{k}\]

    Correct Answer: B

    Solution :

               The equation of a line passing through the points \[A(\mathbf{i}-\mathbf{j}+2\mathbf{k})\] and \[B(3\mathbf{i}+\mathbf{j}+\mathbf{k})\] is                    \[\mathbf{r}=(\mathbf{i}-\mathbf{j}+2\mathbf{k})+\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k})\]                    The position vector of any point P which is a variable point on the line, is \[(\mathbf{i}-\mathbf{j}+2\mathbf{k})+\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k})\]                    \ \[\overrightarrow{AP}=\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k})\Rightarrow |\overrightarrow{AP}|=\lambda \sqrt{11}\]                    Now, if \[\lambda \] \[\sqrt{11}=3\sqrt{11}\] i.e., \[\lambda =3\] then the position vector of the point P is \[10\mathbf{i}+2\mathbf{j}+5\mathbf{k}\].            If \[\lambda \sqrt{11}=-3\sqrt{11},\] i.e., \[\lambda =-3\] then the position vector of the point \[P\] is \[-8\mathbf{i}-4\mathbf{j}-\mathbf{k}\].


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