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Find the vector and cartesian equationof the sphere on the join of the points $A$ and $B$ having position vectors $\overrightarrow{2i}+\overrightarrow{6j}-\overrightarrow{7k}$ and $-\overrightarrow{2i}+\overrightarrow{4j}-\overrightarrow{3k}$ respectively as a diameter . Find also the centre and radius of the sphere.

1 Answer

  • General equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ centre $(-u, -v, -w)$ radius $ = \sqrt{u^2+v^2+w^2-d}$
  • Vector equation of a sphere, the end points of a diameter being $A ( \overrightarrow a)\: and\: B (\overrightarrow b)\: (\overrightarrow r-\overrightarrow a).(\overrightarrow r-\overrightarrow b)=0$ Cartesian form $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+(z-z_1)(z-z_2)=0$
The end points of the diameter have pvs, $ \overrightarrow a = O\overrightarrow A=2\overrightarrow i+6\overrightarrow j-7\overrightarrow k\: and \: \overrightarrow b=O\overrightarrow b=-2\overrightarrow i+4\overrightarrow j-3\overrightarrow k$
The vector equation is $ (\overrightarrow r-\overrightarrow a).(\overrightarrow r-\overrightarrow b)=0$
$(\overrightarrow r-(2\overrightarrow i+6\overrightarrow j-7\overrightarrow k)).(\overrightarrow r-(-2\overrightarrow i+4\overrightarrow j-3\overrightarrow k))=0$
Cartesian equation is
where $(x_1, y_1, z_1)=(2, 6, -7)\: \: (x_2, y_2, z_2)=(-2, 4, -3)$
$ (x-2)(x+2)+(y-6)(y-4)+(z+7)(z+3)=0$
$ x^2+y^2+z^2-10y+10z+41=0$
The centre of the sphere is at $(-u, -v, -w) $ i.e., $(0, 5, -5)$
where $ 2u=0, \: 2v=-10, \: 2w=10$
The radius is $ \sqrt{u^2+v^2+w^2-d} = \sqrt{0+15+25-41} = 3$ units.
answered Jun 18, 2013 by thanvigandhi_1

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