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What is the equation of the line through the origin and the centre of the sphere ?
Concept: 1). The general form of the equation of the sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where u, v, w and d are constant.2). Centre of the sphere is given by (-u, -v, -w)3). Equation of line passing through two points (x1, y1, z1) and (x2, y2, z2) is given by \(\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)Calculation:The given equation of the sphere is x2 + y2 + z2 - 2x - 3y - 4z = 0.Centre of the sphere is given by (-u, -v, -w) ? (1, \(\frac{3}{2}\), 2)Here the two points are (0, 0, 0) and (1, \(\frac{3}{2}\), 2).Equation of line passing through two points (0, 0, 0) and (1, \(\frac{3}{2}\), 2) is given by,\(\displaystyle \frac{x-0}{1-0}=\frac{y-0}{\frac{3}{2}-0}=\frac{z-0}{2-0}\)? x = \(\displaystyle \frac{2y}{3}\) = \(\displaystyle \frac{z}{2}\)? 6x = 4y = 3z? The required equation of the line is 6x = 4y = 3z.
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