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Chapter 9: Problem 20

Find the distance between the points. \((-2,3,2), \quad(2,-5,-2)\)

Short Answer

Expert verified
The distance between points (-2, 3, 2) and (2, -5, -2) is \( \sqrt{96} \) units

Step by step solution

01

Identifying the Coordinates

The given points are (-2, 3, 2) and (2, -5, -2). Here, for the first point, \(x_1 = -2\), \(y_1 = 3\), and \(z_1 = 2\); for the second point, \(x_2 = 2\), \(y_2 = -5\), and \(z_2 = -2\).
02

Apply the Distance Formula

The distance formula in 3-dimensions is \(d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\). So, substituting the respective values, we get \(d=\sqrt{(2-(-2))^2 + ((-5)-3)^2 + ((-2)-2)^2}\)
03

Simplify

Solving the expression inside the brackets, we get \(d=\sqrt{(4)^2 + (-8)^2 + (-4)^2}\) which leads to \(d=\sqrt{16 + 64 + 16}\)
04

Compute Final Answer

On calculating the expression inside the square root, we get \(d= \sqrt{96}\). Hence, the distance is \(\sqrt{96}\) units

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