The Pythagorean theorem can be extended into three dimensions to help find the distance between two points in 3D space. If you have two points, \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), the distance \( d \) between these two points can be calculated using the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Practice Problems
Let's practice with some problems:
Problem 1
Find the distance between the points \( A(1, 2, 3) \) and \( B(4, 6, 8) \).
Solution:
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Calculate the differences in the coordinates:
- \( x_2 - x_1 = 4 - 1 = 3 \)
- \( y_2 - y_1 = 6 - 2 = 4 \)
- \( z_2 - z_1 = 8 - 3 = 5 \)
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Plug these differences into the formula: \[ d = \sqrt{(3)^2 + (4)^2 + (5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]
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Simplify: \[ d = 5\sqrt{2} \]
Problem 2
Calculate the distance between the points \( A(-1, 0, 2) \) and \( B(3, 5, -1) \).
Solution:
-
Calculate the differences in the coordinates:
- \( x_2 - x_1 = 3 - (-1) = 4 \)
- \( y_2 - y_1 = 5 - 0 = 5 \)
- \( z_2 - z_1 = -1 - 2 = -3 \)
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Plug these differences into the formula: \[ d = \sqrt{(4)^2 + (5)^2 + (-3)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} \]
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Simplify: \[ d = 5\sqrt{2} \]
Problem 3
Find the distance between the points \( A(0, 0, 0) \) and \( B(2, 2, 2) \).
Solution:
-
Calculate the differences in the coordinates:
- \( x_2 - x_1 = 2 - 0 = 2 \)
- \( y_2 - y_1 = 2 - 0 = 2 \)
- \( z_2 - z_1 = 2 - 0 = 2 \)
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Plug these differences into the formula: \[ d = \sqrt{(2)^2 + (2)^2 + (2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} \]
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Simplify: \[ d = 2\sqrt{3} \]
Summary
The distance between two points in 3D space can be found using an extension of the Pythagorean theorem. Practice the above problems to become proficient in calculating distances in three-dimensional coordinates! If you’d like more practice or have specific cases, let me know!