Find an equation for the plane consisting of all points that are equidistant from the points (1, 0, -2) and (3, 4, 0) .

Solution

Given, the points are (1, 0, -2) and (3, 4, 0).

We have to find an equation for the plane consisting of all points that are equidistant from the given points.

Let the parametric point be (x, y, z) on the plane which is equidistant from the given points.

Using the distance formula,

\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\)

Distance between the points (x, y, z) and (1, 0, -2) = \(\sqrt{(x-1)^{2}+(y-0)^{2}+(z-(-2))^{2}}\)

= \(\sqrt{(x-1)^{2}+y^{2}+(z+2)^{2}}\)

Distance between the points (x, y, z) and (3, 4, 0) = \(\sqrt{(x-3)^{2}+(y-4)^{2}+(z-0)^{2}}\)

= \(\sqrt{(x-3)^{2}+(y-4)^{2}+z^{2}}\)

Given, \(\sqrt{(x-1)^{2}+y^{2}+(z+2)^{2}}=\sqrt{(x-3)^{2}+(y-4)^{2}+z^{2}}\)

On squaring both sides,

\(\sqrt{(x-1)^{2}+y^{2}+(z+2)^{2}}=\sqrt{(x-3)^{2}+(y-4)^{2}+z^{2}}\)

By using algebraic identity,

(a - b)² = a² - 2ab + b²

(a + b)² = a² + 2ab + b²

(x² - 2x + 1) + y² + (z² + 4z + 4) = (x² - 6x + 9) + (y² - 8y + 16) + z²

x² + y² + z² - 2x + 4z + 1 + 4 = x² + y² + z² - 6x - 8y + 9 + 16

By grouping,

-2x + 6x + 4z - 8y + 5 - 9 - 16 = 0

4x - 8y + 4z - 20 = 0

Dividing by 4 on both sides,

x - 2y + z - 5 = 0

Therefore, the equation of the plane is x - 2y + z - 5 = 0.

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Find an equation for the plane consisting of all points that are equidistant from the points (1, 0, -2) and (3, 4, 0).

Summary:

An equation for the plane consisting of all points that are equidistant from the points (1, 0, -2) and (3, 4, 0) is x - 2y + z - 5 = 0.