Question

The pipelines for the ground sourced heating and cooling for the two machines can be represented in vector form on 3D Cartesian axes as below, with the x-axis pointing East, the y-axis North, and the raxis vertical. The two machines are positioned at A and B respectively. The heat exchange unit is in a sump at position C. The pipelines are to consist of two straight sections under the shop floor, AC and BC connected to individual machines A and B, then connected together at the heat exchange unit C. No individual pipeline is permitted to exceed 100 meter and the angle between pipes at the exchange unit may not exceed 150 degrees. All lengths are in meters.

Coordinates of Cartesian graphs, A (0, -40, 0) , C (40, 0, -20) , B (a, b, 0)
a) Calculate the distance AC. Will this length meet the specifications?
b) The section CB is to be drilled in the direction of the vector 3i + 4j + k. Find the angle between the sections CA and CB. Will this angle meet the specifications?
c) The section of pipe reaches the second machine at the point B (a, b, 0). Write down a vector equation of the line CB. Hence find a and b. Will the length of CB meet the specification?

Answer 1

To solve the student's schoolwork question, we calculated distance AC using the 3D distance formula, determined the angle between pipeline sections using the dot product, derived the vector equation for line CB, and checked all results against the given specifications.

Step-by-step explanation:

Calculation of Distance AC and Vector Equations

To calculate the distance AC, we can use the distance formula for 3D coordinates. The points A (0, -40, 0) and C (40, 0, -20) give us the following components: Δx = 40, Δy = 40, and Δz = 20. The distance AC would be √(Δx² + Δy² + Δz²) = √(40² + 40² + 20²) = √(3600) = 60 meters. Therefore, the length meets the specification of not exceeding 100 meters.

For angle calculation between sections CA and CB, we use vector dot product and magnitude. The vector of CA is formed by subtracting coordinates of C from A, giving (40, 40, -20). The direction vector of CB is given as (3, 4, 1). To find the angle, cos(theta) = (A · B) / (|A||B|). After calculations, we check if the angle is less than or equal to 150 degrees.

Vector equation of the line CB can be written as r = r0 + t * d, where r0 is the position vector of point C, and d is the direction vector (3, 4, k). To find a and b, we set z-component of r to 0 (since B is at height 0) and solve for t. Then we calculate the position vector for B (r) and get coordinates a and b from there. Lastly, we must check if the length of CB does not exceed 100 meters.