A pipeline is to be fitted under a road within the perimeter of the theme park and can be represented on 3D Cartesian axes as below, with the x-axis pointing East, the y-axis North, and the z-axis vertical. The pipeline is to consist of a straight section AB directly under the road, and another straight section BC connected to the first. All lengths are in metres. Refer to the attached image and provide the following: 1- Calculate the distance AB The section BC is to be drilled in the direction of the vector 3i+4j+k 2-Find the angle between the sections AB and BC. The section of pipe reaches ground level at the point (a,b,0). 3-Write down a vector equation of the line BC. Hence find a and b.
A pipeline is to be fitted under a road within the perimeter of the theme park and can be represented on 3D Cartesian axes as below, with the x-axis pointing East, the y-axis North, and the z-axis vertical. The pipeline is to consist of a straight section AB directly under the road, and another straight section BC connected to the first. All lengths are in metres.
Refer to the attached image and provide the following:
1- Calculate the distance AB
The section BC is to be drilled in the direction of the
2-Find the angle between the sections AB and BC.
The section of pipe reaches ground level at the point (a,b,0).
3-Write down a vector equation of the line BC. Hence find a and b.
![A (0, -40, 0)
y (N)
C (a, b, 0)
B (40, 0, -20)
x (E)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc32c1e5f-b2a4-418c-8d03-158c2a8cd1bc%2Fc636a026-64f5-41d6-96ba-00e3be7c1f59%2F2oc43c_processed.png&w=3840&q=75)
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Hi, Unfortunately when i view the answer on my laptop it shows in text form as shown in the attached image.
It displays correctly on a phone or laptop, however it cuts of the right hand side of the equations.
Is there a way to fix this?
![Step 3
Since, 40,0,-20is on the section BC.
It must satisfy the equation of line,
..40=3t+a,0=4t+b,-20=t.Therefore,t=-20,a=40-3-
20a=40+60a 100and b=-4tb=-4×-20b-80..Point a,b,0=100,80,0BC=100-40i+80-
Oj+0+20KBC-=60i+80j+20kAlso,AB¬=40-0i+0+40j+-20-0kAB¬=40i+40j-20k.
Step 4
Hence,angle between AB and BC
=cos-1AB .BC-AB-BC-=cos-12400+3200-4003600+6400+4001600+1600+400=cos-
1520010104x60=cos-1526104-cos-1521046x104-cos-110412
Therefore, the angle between AB and BC =cos-110412.
(3) From part(2) equation of line BC is
x-1003-y-804=Z-01
and
a=100, b=80](https://content.bartleby.com/qna-images/question/c32c1e5f-b2a4-418c-8d03-158c2a8cd1bc/403fd667-3d60-49f1-9eab-a03ad41b2163/ehakzyb_thumbnail.png)
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