What Is the Triangle Law of Vector Addition?

The triangle law of vector addition establishes a geometric method for obtaining the resultant of two vectors by constructing a triangle whose sides represent their magnitudes and directions.

Formal Statement and Diagrammatic Representation of the Triangle Law of Vector Addition

Definition: If two vectors are represented in magnitude and direction by two sides of a triangle taken in order, their resultant is represented by the third side of the triangle taken in the reverse order. If $\vec{A}$ and $\vec{B}$ are the two vectors, then the resultant $\vec{R}$ is given by $\vec{R} = \vec{A} + \vec{B}$. The tail of $\vec{B}$ is placed at the head of $\vec{A}$, and $\vec{R}$ is drawn from the tail of $\vec{A}$ to the head of $\vec{B}$.

In this construction, $\vec{A}$ and $\vec{B}$ must be arranged such that their directions and magnitudes are preserved, maintaining vector equivalence through translation. For foundational definitions and operations, see Introduction to Vector Algebra.

Standard Formulae for Magnitude and Direction of the Resultant

Let vectors $\vec{A}$ and $\vec{B}$ make an angle $\theta$ with each other. The magnitude of the resultant vector $\vec{R}$ by the triangle law is

$|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$

The direction of $\vec{R}$, measured as angle $\phi$ from $\vec{A}$, is given by

$\phi = \tan^{-1} \left( \dfrac{B\sin\theta}{A+B\cos\theta} \right)$

These equations remain valid for all pairs of non-zero vectors, provided the included angle $\theta$ is well-defined and measured from $\vec{A}$ towards $\vec{B}$.

Derivation of Resultant Expression Using Triangle Construction

Consider $\vec{A}$ represented by segment $OA$ and $\vec{B}$ by segment $AB$, forming triangle $OAB$. Extend $OA$ to $C$ such that $BC$ is perpendicular to $OC$, and let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$.

In $\triangle OBC$ (right-angled at $B$), apply the Pythagorean theorem:

$OB^2 = OC^2 + BC^2$

Express $OC$ and $BC$ in terms of $A$, $B$, and $\theta$ using triangle $ABC$:

$AC = B\cos\theta$, $BC = B\sin\theta$, $OC = OA + AC = A + B\cos\theta$

Substitute into the Pythagorean identity:

$\begin{aligned} R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2 \end{aligned}$

Expand and simplify step by step:

$\begin{aligned} R^2 &= A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta \\ &= A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta) \\ &= A^2 + 2AB\cos\theta + B^2 \end{aligned}$

Thus, $R = \sqrt{A^2 + 2AB\cos\theta + B^2}$ as required. Result: The formula expresses the resultant’s magnitude when two vectors are joined sequentially.

For the direction $\phi$:

$\tan\phi = \dfrac{BC}{OC} = \dfrac{B\sin\theta}{A+B\cos\theta}$

Therefore, $\phi = \tan^{-1} \left( \dfrac{B\sin\theta}{A+B\cos\theta} \right)$ gives the angle between $\vec{A}$ and $\vec{R}$.

Specific Cases and Geometric Consequences

If $\theta = 0^\circ$, the vectors are collinear and point in the same direction; $|\vec{R}| = A + B$.

If $\theta = 180^\circ$, the vectors are opposite; $|\vec{R}| = |A - B|$.

If $\theta = 90^\circ$, the vectors are perpendicular; $|\vec{R}| = \sqrt{A^2 + B^2}$.

Difference Between Vectors and Scalars clarifies relevant distinctions for these limiting scenarios.

Illustrative Problems Based on the Triangle Law of Vector Addition

Example: Two vectors, $\vec{P}$ and $\vec{Q}$, have magnitudes $6$ units and $8$ units, respectively. The angle between them is $60^{\circ}$. Find the magnitude and direction of their resultant.

Solution: Given $P = 6$, $Q = 8$, $\theta = 60^\circ$.

$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$

$= \sqrt{6^2 + 8^2 + 2 \times 6 \times 8 \times 0.5}$

$= \sqrt{36 + 64 + 48}$

$= \sqrt{148}$

$= 12.165$ units

Direction:

$\phi = \tan^{-1} \left( \dfrac{8 \times \sin 60^\circ}{6 + 8 \times \cos 60^\circ} \right)$

$= \tan^{-1} \left( \dfrac{8 \times 0.866}{6 + 8 \times 0.5} \right)$

$= \tan^{-1} \left( \dfrac{6.928}{10} \right)$

$= \tan^{-1} (0.6928)$

$= 34.7^\circ$

Result: The resultant’s magnitude is $12.165$ units and it is at $34.7^\circ$ with respect to $\vec{P}$.

Example: If the magnitudes of two vectors are $9$ units and $4$ units and the resultant magnitude is $11$ units, find the angle between the vectors.

Solution: Let $A=9$, $B=4$, $R=11$.

$R^2 = A^2 + B^2 + 2AB\cos\theta$

$121 = 81 + 16 + 72\cos\theta$

$121 - 97 = 72\cos\theta$

$24 = 72\cos\theta$

$\cos\theta = \dfrac{24}{72} = \dfrac{1}{3}$

$\theta = \cos^{-1}(\dfrac{1}{3}) \approx 70.53^\circ$

Result: The vectors are approximately at $70.53^\circ$ to each other.

Example: Two vectors, each of magnitude $a$, are perpendicular. Determine the resultant’s magnitude and orientation with respect to the first vector.

Solution: $A = a$, $B = a$, $\theta = 90^\circ$.

$R = \sqrt{a^2 + a^2 + 2a^2 \times 0} = \sqrt{2}a$

$\phi = \tan^{-1} \left( \dfrac{a}{a} \right) = \tan^{-1}(1) = 45^\circ$

Result: The resultant has magnitude $\sqrt{2}a$ and direction $45^\circ$ from the first vector.

Example: Calculate the resultant of vectors $\vec{A}$ and $\vec{B}$ both of magnitude $5$ units, but opposite in direction.

Solution: $A=5$, $B=5$, $\theta = 180^\circ$.

$R = \sqrt{5^2 + 5^2 + 2\times 5 \times 5 \times (-1)}$

$= \sqrt{25 + 25 - 50}$

$= \sqrt{0} = 0$

Result: The resultant vector is the zero vector when two equal and opposite vectors are added.

For more graphical methods of vector addition or advanced examples, see Graphical Methods of Vector Addition.

Exam Cues and Common Misconceptions Related to the Triangle Law of Vector Addition

• Vectors must be translated, not rotated, for head-to-tail arrangement
• Angle $\theta$ must be properly defined between vectors
• Direction formula gives angle with respect to the first vector always
• Triangle law is exclusive to the addition of two vectors
• Graphical and algebraic forms must yield the same resultant

Component form and parallelogram law interpretations are mathematically equivalent to the triangle method for two vectors; see Understanding Vector Triple Product for multi-vector product operations.