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

2910 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Particles P and Q move in a plane with constant velocities. At time t = 0 the position vectors of P and Q, relative to a fixed point O in the plane, are (16i - 12j) m and -5i + 4j) m respectively. The velocity of P is (i + 2j) m/s and the velocity of Q


In statistics, what is the benefit of taking a sample survey rather than a census?


Prove ∑r^3 = 1/4 n^2(n+1)^2


How do i figure out if integrals are improper or not and how do i know which limit is undefined?


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:

© 2026 by IXL Learning