Adding Vectors Geometrically — Question 6

Question 6

Magnitude bounds can reject an impossible drawing

Two displacement vectors have fixed magnitudes 7 m and 3 m, but their relative direction is adjustable in a plane. A proposed construction is acceptable only if the arrows are translated without changing their lengths. Three students claim resultant magnitudes of 2 m, 4 m and 9 m.

1. Derive the full attainable interval of resultant magnitudes using triangle geometry. Classify each proposed magnitude.

2. Describe the configurations giving the smallest and largest magnitudes. Is the smaller vector’s direction also the minimum resultant’s direction?

3. Construct the 9 m case using a triangle with side lengths 7, 3 and 9 m. Explain why two mirror-image joins are possible when the target arrow is fixed eastward.

4. If a third freely directed vector of length 5 m is added, can the three arrows close? Give a geometric existence argument, and compare the zero resultant with their total route length.

Question 6 - Solution

Magnitude bounds can reject an impossible drawing

(a) Apply both triangle inequalities, including degenerate triangles.

$$|7-3|\le R\le7+3\Rightarrow\boxed{4\le R\le10\quad\mathrm m};\quad2\notin[4,10],\quad4,9\in[4,10]$$ The upper bound follows from the triangle inequality; the lower bound follows because the shorter arrow cannot cancel more than 3 m of the longer one. Continuous angle variation attains all intermediate lengths.

(b) Alignment fixes the extreme cases.

$$R_{\min}=7-3=4\,\mathrm m,\quad R_{\max}=7+3=10\,\mathrm m$$ The minimum uses opposite arrows and points along the 7 m arrow. The maximum uses parallel arrows in the same direction.

(c) Use distances from the two target endpoints.

$$OP=7,\quad PT=3,\quad OT=9\quad\mathrm m,\qquad|7-3|<9<7+3$$ The strict inequalities make the two construction circles intersect twice, giving mirror-image joins above and below the eastward target. Both are valid but use different ordered vectors.

(d) A closing arrow must oppose the first pair’s resultant.

$$5\in[4,10]\Rightarrow\vec d_3=-(\vec d_1+\vec d_2),\quad\sum\vec d_i=\vec0,\quad s=7+3+5=15\,\mathrm m$$ First arrange the 7 m and 3 m arrows to give a 5 m resultant, then place the third arrow opposite to it. Closure makes displacement zero without making route length zero.