Adding Vectors Geometrically — Question 5

Question 5

Two possible constructions can satisfy the same target

A resultant displacement $\overrightarrow{OT}$ is fixed at 8 m east. It must be made from two successive displacement vectors, each of magnitude 5 m. If P is the join, then OP and PT must both be 5 m. The motion is in a horizontal plane with north upward.

1. Describe a compass construction locating every possible P. How many solutions are there, and what are their positions relative to the midpoint of OT?

2. Determine the directions of both legs in each solution using right-triangle geometry. Distinguish the ordered pair of arrows from its resultant.

3. If an extra rule requires the first leg to point north of east, which solution survives?

4. Replace the target length 8 m by a nonnegative length $r$, keeping both legs 5 m. Classify the constructions for $0<r<10$ m, $r=10$ m, $r>10$ m and $r=0$. Explain the special zero-target case.

Question 5 - Solution

Two possible constructions can satisfy the same target

(a) A valid join must satisfy two distance constraints at once.

$$OP=PT=5\,\mathrm m,\quad OM=4\,\mathrm m\Rightarrow\boxed{MP=\sqrt{25-16}=3\,\mathrm m}$$ Intersect radius-5 m circles centered at O and T. Their two intersections lie 3 m north and south of midpoint M, giving both possible joins.

(b) The direction choices are paired, not independent.

$$\alpha=\tan^{-1}(3/4)=36.870^\circ,\quad(\theta_1,\theta_2)=(+\alpha,-\alpha)\text{ or }(-\alpha,+\alpha)$$ The signs are paired: an upward first leg needs a downward second leg, or vice versa. Both paths have 10 m traveled distance and 8 m displacement east.

(c) Additional directional information selects one branch.

$$P_y>0\Rightarrow P=(4,3)\,\mathrm m,\quad\vec{PT}=(4,-3)\,\mathrm m$$ The additional directional condition selects the northern intersection. Choosing both legs north of east would miss the target.

(d) Count circle intersections, including degeneracies.

$$N(r)=\begin{cases}2&0<r<10\,\mathrm m,\\1&r=10\,\mathrm m,\\0&r>10\,\mathrm m,\\\infty&r=0.\end{cases}$$ Two crossing circles give two joins; tangency gives one; separated circles give none. At zero target they coincide, allowing every first direction followed by its opposite. The zero resultant has no eastward heading.