Triangle Law for Vector Addition

Worked out. Given a triangle with vertices A, B, and C (shown in a small accompanying diagram), the problem asks you to express the vector $\vec{AC}$ in terms of the two given vectors $\vec{AB}$ and $\vec{CB}$. The method applies the triangle law of vector addition to the closed path A to C to B, writing $\vec{AC}+\vec{CB}=\vec{AB}$ and rearranging to isolate $\vec{AC} = \vec{AB}-\vec{CB}$ — a direct application of the triangle law together with the fact that a vector reversed in direction becom …

Worked out. Given a figure showing four force vectors $\vec A_1, \vec A_2, \vec A_3, \vec A_4$ drawn head-to-tail from point O through points A, B, C to D, the problem asks for their combined resultant. The method applies the triangle law REPEATEDLY: first $\vec{OB}=\vec{OA}+\vec{AB}=\vec A_1+\vec A_2$, then $\vec{OC}=\vec{OB}+\vec{BC}=\vec A_1+\vec A_2+\vec A_3$, and finally $\vec{OD}=\vec{OC}+\vec{CD}=\vec A_1+\vec A_2+\vec A_3+\vec A_4$ — showing that the single vector $\vec{OD}$, drawn straight from the starting point O to the final point D, is the resultant of chaining several vectors he …

What this figure shows. Two-panel figure. Panel (a) shows two vectors $\vec A$ and $\vec B$ drawn in a plane, not along the same line, both starting from roughly the same region but at an angle to each other (their relative orientation as originally drawn, before addition). Panel (b) shows the SAME two vectors redrawn head-to-tail — $\vec B$'s tail placed at $\vec A$'s head — with a third arrow labelled $\vec C$ (marked 'Resultant') drawn from the tail of $\vec A$ to the head of $\vec B$, closing the triangle and representing $\vec C=\vec A+\vec B$ by the tri …

What this figure shows. A single diagram with a common origin O and two different triangles sharing the same final point B, illustrating that vector addition is commutative. Triangle O-A-B shows $\vec P$ drawn from O to A, then $\vec Q$ drawn from A to B, giving resultant $\vec R=\vec{OB}$ (representing $\vec P+\vec Q$). Triangle O-C-B shows the SAME two vectors added in the opposite order — $\vec Q$ drawn from O to C, then $\vec P$ drawn from C to B — reaching the SAME final point B, so $\vec Q+\vec P$ also equals $\vec{OB}=\vec R$. Both routes give the identical resultant, visually proving $\vec P+\vec Q=\vec Q+\vec P$. Points O, …

What this figure shows. A single diagram with points O, P, Q, R showing three vectors $\vec A$, $\vec B$, $\vec C$ added in two different groupings that both reach the same final resultant $\vec R$. One path (through point Q) first combines $\vec A+\vec B$ and then adds $\vec C$ to reach R, illustrating $(\vec A+\vec B)+\vec C$ via triangle O-Q-R. The other path (through point P) first combines $\vec B+\vec C$ and then adds $\vec A$ to reach the SAME point R, illustrating $\vec A+(\vec B+\vec C)$ via triangle O-P-R. Both groupings arrive at the identical resultant vector $\vec R=\vec{OR}$, visually proving the associative law $(\vec A+\vec B)+\vec C=\vec A …