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Find the shortest distance between the lines r = (1, 5, 6) + y(-2, -1, 0) and r = (1, 7, -3) + z(2, 0, 4)

Vector joining the two lines = (1, 5, 6) - (1, 7, -3) = (0, -2, 9)Normal vector to the two lines = (-2, -1, 0) x (2, 0, 4) = (-4, 8, 2) = 2(-2, 4, 1)Hence, using the dot product, shortest distance = (0, -2, 9) "dot" (-2, 4, 1) / sqrt(2² + 4² + 1²) = -8 + 9 / sqrt(4 + 16 + 1) = 1/sqrt(21)

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Find the shortest distance between the lines r = (1, 5, 6) + y(-2, -1, 0) and r = (1, 7, -3) + z(2, 0, 4)

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Find the shortest distance between the lines r = (1, 5, 6) + y(-2, -1, 0) and r = (1, 7, -3) + z(2, 0, 4)

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Prove by induction that 6^n + 4 is divisible by 5 for all integers n >= 1

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We're here to help

Company Information

Popular Requests

© MyTutorWeb Ltd 2013–2025

CLICK CEOP

The infinite series C and S are defined C = acos(x) + a^2cos(2x) + a^3cos(3x) + ..., and S = asin(x) + a^2sin(2x) + a^3sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = asin(x)/(1 - 2acos(x) + a^2), and find C.