What is the distance from point P_{1}=(2,-2,3) quad to P_{2}=(-3,2,-1) ?

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Step:1 ：We are given two points in three-dimensional space, P_{1}=(2,-2,3) and P_{2}=(-3,2,-1), and we are asked to find the distance between these two points.Step:2 ：The distance between two points in three-dimensional space can be calculated using the Euclidean distance formula, which is given by: [d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}]Step:3 ：Substituting the given coordinates into the formula, we get: [d = sqrt{(-3-2)^2 + (2-(-2))^2 + (-1-3)^2}]Step:4 ：Solving the above expression, we get the distance as approximately 7.55.Step:5 ：(boxed{7.55}) is the final answer.

Answer

Step: 1 ：We are given two points in three-dimensional space, P_{1}=(2,-2,3) and P_{2}=(-3,2,-1), and we are asked to find the distance between these two points.Step: 2 ：The distance between two points in three-dimensional space can be calculated using the Euclidean distance formula, which is given by: [d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}]Step: 3 ：Substituting the given coordinates into the formula, we get: [d = sqrt{(-3-2)^2 + (2-(-2))^2 + (-1-3)^2}]Step: 4 ：Solving the above expression, we get the distance as approximately 7.55.Step: 5 ：(boxed{7.55}) is the final answer.

What is the distance from point $P_{1}=(2,-2,3) \quad$ to $P_{2}=(-3,2,-1)$ ?

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