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
Answered by Abhinav H. Further Mathematics tutor

2410 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Find arsinh(x) in terms of x


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.


Integrate (4x+3)^1/2 with respect to x.


Express (X²-16)/(X-1)(X+3) in partial fractions


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning