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)

AH

Related Further Mathematics A Level answers

All answers ▸

Can you express 3 + 4j in polar form?


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


Use the geometric series e^(ix) - (1/2)e^(3ix) + (1/4)e^(5ix) - ... to find the exact value sin1 -(1/2)sin3 + (1/4)sin5 - ...


Find all the cube roots of 1