Cable Displacement Vectors Solution

In structural construction, cable lines are installed to anchor tall towers. Construction managers must determine the length and spatial direction of the cables from the coordinates of the anchor points to ensure proper procurement and site layout.

A support tower has its base at the origin and rises vertically along the $z$-axis. A cable is to be connected from a point $P(0, 0, 45)\text{ meters}$ high on the tower to a ground anchor situated at $Q(15, -20, 0)\text{ meters}$.

1. Find the displacement vector $\vec{PQ}$ representing the cable in component form.
2. Find the total length of the cable required (the magnitude of the vector).
3. Find the unit vector in the direction of the cable.

Step-by-Step Solution:

1. Find the Displacement Vector $\vec{PQ}$: The displacement vector is determined by subtracting the coordinates of the initial point $P(0,0,45)$ from the terminal point $Q(15,-20,0)$:

   $$\vec{PQ} = \langle 15-0, -20-0, 0-45 \rangle = \langle 15, -20, -45 \rangle\text{ meters}$$

2. Find the Cable Length (Magnitude): The length of the cable is the magnitude of the displacement vector:

   $$|\vec{PQ}| = \sqrt{15^2 + (-20)^2 + (-45)^2} = \sqrt{225 + 400 + 2025} = \sqrt{2650}$$
   $$|\vec{PQ}| = 5\sqrt{106} \approx 51.48\text{ meters}$$

3. Find the Unit Vector ($\hat{U}$): Divide the displacement vector by its magnitude:

   $$\hat{U} = \frac{\vec{PQ}}{|\vec{PQ}|} = \frac{\langle 15, -20, -45 \rangle}{5\sqrt{106}} = \left\langle \frac{3}{\sqrt{106}}, -\frac{4}{\sqrt{106}}, -\frac{9}{\sqrt{106}} \right\rangle$$
   Evaluating the components:
   $$\hat{U} \approx \langle 0.291, -0.388, -0.874 \rangle$$

Meaning of the Answer & Real-Life Application:

Calculating displacement vectors and magnitudes allows construction managers to order the correct length of steel cables. The unit vector represents the direction of the tension force that the cable will exert on both the tower and the ground anchor, which is essential for structural stability calculations.