Can anyone help me with this vector question?

A pipeline is to be fitted beneath a road as represented in the image. All lengths are in meters.

A) Calculate the length of the pipe AB. Use diagrams to support your answer where required.
B) The section BC is to be drilled in the direction 3i + 4j + k. Find the angle between the sections AB and BC.
C) The pipe ends at point C (a,b,0). Give a vector equation of the line BC. Hence find a and b.

enter image source here

1 Answer
Mar 24, 2018

See below.

Explanation:

Given

A = (0,-20,0)A=(0,20,0)
B = (20,0,-10)B=(20,0,10)
C = (a,b,0)C=(a,b,0)

we have

bar(AB) = norm(A-B) = sqrt(20^2+20^2+10^2)¯¯¯¯¯¯AB=AB=202+202+102

Segment s_1 = A+lambda (B-A)s1=A+λ(BA) with lambda in [0,1]λ[0,1] has the direction of B-A = (20,20,-10)BA=(20,20,10)

now with vec n = (3,4,1) = mu (C-B)n=(3,4,1)=μ(CB) we have

<< B-A, vec n >> = norm(B-A) norm(vec n) cos phiBA,n=BAncosϕ and then

cos phi = ( << B-A, vec n >> )/( norm(B-A) norm(vec n)) = (20 xx 3+20 xx 4-10 xx 1)/(sqrt(20^2+20^2+10^2)sqrt(3^2+4^2+1))cosϕ=BA,nBAn=20×3+20×410×1202+202+10232+42+1

and finally

C = B + rho vec nC=B+ρn or

((a),(b),(0)) = ((20),(0),(-10)) + rho((3),(4),(1))

and solving we get

rho = 10, b = 40, a = 50