JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    The radius of the circular section of the sphere \[|\mathbf{r}|\,=5\]by the plane \[\mathbf{r}\,.\,(\mathbf{i}+\mathbf{j}+\mathbf{k})=3\sqrt{3}\] is [DCE 1999]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: D

    Solution :

    • The centre of the sphere \[|r|\,=5\] is at the origin and radius\[=5\]. Let M be the foot of perpendicular from O to the given plane.  Then OM = length of perpendicular from O to the given plane \[=\frac{|\overrightarrow{OM}\,.\,(i+j+k)-3\sqrt{3}|}{|i+j+k|}\]
    • \[=\frac{3\sqrt{3}}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}}=3\]                   
    • Let P be any position of circle, then P lies on plane as well as on sphere.                   
    • \[\therefore \] OP = radius of sphere = 5                   
    • In \[\Delta OPM\], we have \[O{{P}^{2}}=O{{M}^{2}}+P{{M}^{2}}\]                   
    • \[\Rightarrow PM=\sqrt{{{5}^{2}}-{{3}^{2}}}=4\].

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