The Pythagorean Theorem can be extended into three dimensions to calculate the distance between two points in 3D space. If you have two points in three-dimensional space, \(A(x_1, y_1, z_1)\) 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
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Problem 1: Calculate the distance between points \(A(1, 2, 3)\) and \(B(4, 6, 8)\).
- Solution: \[ d = \sqrt{(4 - 1)^2 + (6 - 2)^2 + (8 - 3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \approx 7.07 \]
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Problem 2: Find the distance between points \(C(0, 0, 0)\) and \(D(1, 1, 1)\).
- Solution: \[ d = \sqrt{(1 - 0)^2 + (1 - 0)^2 + (1 - 0)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73 \]
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Problem 3: Determine the distance between points \(E(-1, -1, -1)\) and \(F(2, 3, 4)\).
- Solution: \[ d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2 + (4 - (-1))^2} = \sqrt{(3)^2 + (4)^2 + (5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \approx 7.07 \]
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Problem 4: Calculate the distance between points \(G(3, 4, 5)\) and \(H(6, 8, 10)\).
- Solution: \[ d = \sqrt{(6 - 3)^2 + (8 - 4)^2 + (10 - 5)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \approx 7.07 \]
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Problem 5: Find the distance between points \(I(-2, 3, 7)\) and \(J(0, -1, 1)\).
- Solution: \[ d = \sqrt{(0 - (-2))^2 + (-1 - 3)^2 + (1 - 7)^2} = \sqrt{(2)^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} \approx 7.48 \]
Additional Practice
Try calculating the distance for the following pairs of points:
- \(K(1, 2, 2)\) and \(L(3, 5, 5)\)
- \(M(-3, -3, 0)\) and \(N(2, 1, 1)\)
Summary
The extension of the Pythagorean theorem into three dimensions allows us to find distances in three-dimensional space effectively using coordinates. The formula involves calculating the square of the differences in each dimension (x, y, and z), summing them, and then taking the square root of that sum.
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