Question:medium

A man needs to hang two lanterns on a straight wire whose end points have coordinates A (4, 1, -2) and B (6, 2, -3). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.

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Quick Tip: To trisect a line segment, use the section formula. The formula allows you to divide a line in a given ratio, which is useful when dividing a segment into equal parts.
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Solution and Explanation

Given points \( A(4, 1, -2) \) and \( B(6, 2, -3) \). The wire connecting these points is trisected by points \( P_1 \) and \( P_2 \). \( P_1 \) is closer to A and divides the segment in a 1:2 ratio, while \( P_2 \) is closer to B and divides the segment in a 2:1 ratio. 1. Coordinates of \( P_1 \): Using the section formula for a point dividing a line segment \( (x_1, y_1, z_1) \) to \( (x_2, y_2, z_2) \) in the ratio \( m:n \): \[ x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}, \quad z = \frac{m z_2 + n z_1}{m + n} \] For \( P_1 \), \( m=1 \) and \( n=2 \): \[ x_{P_1} = \frac{1 \cdot 6 + 2 \cdot 4}{1 + 2} = \frac{6 + 8}{3} = \frac{14}{3} \] \[ y_{P_1} = \frac{1 \cdot 2 + 2 \cdot 1}{1 + 2} = \frac{2 + 2}{3} = \frac{4}{3} \] \[ z_{P_1} = \frac{1 \cdot (-3) + 2 \cdot (-2)}{1 + 2} = \frac{-3 - 4}{3} = \frac{-7}{3} \] Thus, \( P_1 = \left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right) \). 2. Coordinates of \( P_2 \): For \( P_2 \), \( m=2 \) and \( n=1 \): \[ x_{P_2} = \frac{2 \cdot 6 + 1 \cdot 4}{2 + 1} = \frac{12 + 4}{3} = \frac{16}{3} \] \[ y_{P_2} = \frac{2 \cdot 2 + 1 \cdot 1}{2 + 1} = \frac{4 + 1}{3} = \frac{5}{3} \] \[ z_{P_2} = \frac{2 \cdot (-3) + 1 \cdot (-2)}{2 + 1} = \frac{-6 - 2}{3} = \frac{-8}{3} \] Thus, \( P_2 = \left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right) \). The coordinates of the lantern points are \( P_1 \left( \frac{14}{3}, \frac{4}{3}, \frac{-7}{3} \right) \) and \( P_2 \left( \frac{16}{3}, \frac{5}{3}, \frac{-8}{3} \right) \).
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