Adding Vectors Geometrically — Question 2

Question 2

Choose the correct diagonal of a parallelogram

Two displacement arrows share a tail O. Arrow $\vec A$ is 6 m east; arrow $\vec B$ is 6 m north. Their heads are P and Q respectively. Complete the square OPRQ, where R is opposite O. All four sides have length 6 m.

1. Identify the directed diagonal representing $\vec A+\vec B$. Find its magnitude and direction.

2. Explain how the two routes O–P–R and O–Q–R illustrate that exchanging the order of addition leaves the sum unchanged.

3. Is the directed diagonal P–Q also the sum? Identify the vector equation it satisfies and compare its magnitude and direction with the sum.

4. Both diagonals happen to have equal lengths. Explain why this does not make them equal vectors, and whether equal diagonal lengths are guaranteed for arbitrary parallelograms.

Question 2 - Solution

Choose the correct diagonal of a parallelogram

(a) Start at the common tail, not at one of the heads.

$$\overrightarrow{OR}=\vec A+\vec B,\quad\boxed{R=\sqrt{6^2+6^2}=6\sqrt2\,\mathrm m},\quad\theta=45^\circ$$ The correct sum is the common-tail diagonal O–R, directed northeast. The other diagonal starts at a head, so it represents a different operation.

(b) The two translated-arrow routes reach the same corner.

$$\overrightarrow{OP}+\overrightarrow{PR}=\overrightarrow{OQ}+\overrightarrow{QR}=\overrightarrow{OR}$$ The two routes reach R via different intermediate vertices. This gives the geometric commutative law without claiming the journeys are identical.

(c) The head-to-head diagonal completes a different equation.

$$\vec A+\overrightarrow{PQ}=\vec B\Rightarrow\overrightarrow{PQ}=\vec B-\vec A,\quad|PQ|=6\sqrt2\,\mathrm m,\quad\theta_{PQ}=135^\circ$$ P–Q has the same length as O–R but points northwest. Reversing P–Q would give A minus B instead.

(d) Length alone is insufficient, and the square is a special case.

$$|\vec A+\vec B|^2-|\vec A-\vec B|^2=4AB\cos\phi=0\quad\Leftrightarrow\quad\phi=90^\circ\quad(A,B>0)$$ For nonzero sides, equal diagonal lengths require perpendicular generating arrows. A general parallelogram need not meet that condition. Even in this square, different directions prevent vector equality.