# Triangle Law of Vector Addition

## What is Vector Addition?

Triangle law of vector addition is one of the vector addition laws. Vector addition is defined as the geometrical sum of two or more vectors as they do not follow regular laws of algebra. The resultant vector is known as the composition of a vector.

There are a few conditions that are applicable for any vector addition, they are:

• Scalars and vectors can never be added.
• For any two scalars to be added, they must be of the same nature. Example, mass should be added with mass and not with time.
• For any two vectors to be added, they must be of the same nature. Example, velocity should be added with velocity and not with force.

There are two laws of vector addition, they are:

• Triangle law of vector addition
• Parallelogram law of vector addition

## What is Triangle Law of Vector Addition?

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

To obtain $\vec{R}$ which is the resultant of the sum of vectors $\vec{A}$ and $\vec{B}$ with the same order of magnitude and direction as shown in the figure, we use the following rule:

 $\vec{R}=\vec{A}+\vec{B}$

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## Triangle Law of Vector Addition Derivation

Consider two vectors $\vec{P}$ and $\vec{Q}$ that are represented in the order of magnitude and direction by the sides OA and AB, respectively of the triangle OAB. Let $\vec{R}$be the resultant of vectors $\vec{P}$ and $\vec{Q}$.

From triangle OCB,

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

$OB^{2}=(OA+AC)^{2}+BC^{2}$ (eq.1)

In triangle ACB with ϴ as the angle between P and Q

$cos\Theta =\frac{AC}{AB}$

$AC=ABcos\Theta =Qcos\Theta$

$sin\Theta =\frac{BC}{AB}$

$BC=ABsin\Theta =Qsin\Theta$

Substituting the values of AC and BC in (eqn.1), we get

$R^{2}=(P+Qcos\Theta )^{2}+(Qsin\Theta )^{2}$

$R^{2}=P^{2}+2PQcos\Theta +Q^{2}cos^{2}\Theta +Q^{2}sin^{2}\Theta$

$R^{2}=P^{2}+2PQcos\Theta +Q^{2}$

therefore, $R=\sqrt{P^{2}+2PQcos\Theta +Q^{2}}$

Above equation is the magnitude of the resultant vector.

To determine the direction of the resultant vector, let ɸ be the angle between the resultant vector R and P.

From triangle OBC,

$tan\phi =\frac{BC}{OC}=\frac{BC}{OA+AC}$

$tan\phi =\frac{Qsin\Theta }{P+Qcos\Theta }$

therefore, $\phi =tan^{-1}(\frac{Qsin\Theta }{P+Qcos\Theta })$

Above equation in the direction of the resultant vector.

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