The least distance of the plane 2x + y - 2z - | vedclass

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(D) The given equation of the sphere is $x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0$.Completing the squares,we get $(x - 1)^2 + (y + 2)^2 + (z - 3)^2 = -

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$2$

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(D) The given equation of the sphere is $x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0$.Completing the squares,we get $(x - 1)^2 + (y + 2)^2 + (z - 3)^2 = -10 + 1 + 4 + 9 = 4$.Thus,the center of the sphere is $C(1, -2, 3)$ and the radius $r = \sqrt{4} = 2$.Now,find the perpendicular distance $d$ from the center $C(1, -2, 3)$ to the plane $2x + y - 2z - 6 = 0$ using the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|2(1) + 1(-2) - 2(3) - 6|}{\sqrt{2^2 + 1^2 + (-2)^2}} = \frac{|2 - 2 - 6 - 6|}{\sqrt{4 + 1 + 4}} = \frac{|-12|}{\sqrt{9}} = \frac{12}{3} = 4$.The least distance between the sphere and the plane is given by $d - r = 4 - 2 = 2$ units.

The least distance of the plane $2x + y - 2z - 6 = 0$ from the sphere $x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0$ is .......... $unit$.

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