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

  • question_answer
    The vector equation of a plane, which is at a distance of 8 unit from the origin and which is normal to the vector \[2\mathbf{i}+\mathbf{j}+2\mathbf{k},\] is

    A)            \[\mathbf{r}.(2\mathbf{i}+\mathbf{j}+\mathbf{k})=24\]

    B)            \[\mathbf{r}.(2\mathbf{i}+\mathbf{j}+2\mathbf{k})=24\]

    C)            \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=24\]

    D)            None of these

    Correct Answer: B

    Solution :

               Here \[d=8\] and \[\mathbf{n}=2\mathbf{i}+\mathbf{j}+2\mathbf{k})\]                    \ \[\mathbf{\hat{n}}=\frac{\mathbf{n}}{|\mathbf{n}|}=\frac{2\mathbf{i}+\mathbf{j}+2\mathbf{k}}{\sqrt{4+1+4}}=\frac{2}{3}\mathbf{i}+\frac{1}{3}\mathbf{j}+\frac{2}{3}\mathbf{k}\]                    Hence, the required equation of the plane is                    \[\mathbf{r}.\left( \frac{2}{3}\mathbf{i}+\frac{1}{3}\mathbf{j}+\frac{2}{3}\mathbf{k} \right)=8\] or \[\mathbf{r}.(2\mathbf{i}+\mathbf{j}+2\mathbf{k})=24\].


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