Triangle Law of Vector Addition, Derivation and Examples for JEE

A vector is a quantity or an object that has both magnitude and direction as attributes. Both of these features are required to understand a vector completely. A scalar, on the other hand, is a quantity with only magnitude and no direction. In physics, vectors and scalars are significant. Displacement is a fantastic example of a vector quantity. Displacement indicates our distance from a given point as well as our orientation in relation to that point. Distance is an example of a scalar and shows us how far away we are from a fixed location, but it doesn't tell us how to get there.

The addition of scalars can be done algebraically because vectors are not scalars and do not follow the laws of scalar algebra. This is due to the fact that vectors have both magnitude and direction. Geometrically, vectors are added. The addition or composition of vectors occurs when two or more vectors are combined. When two or more vectors are joined together, the resultant vector is the result.

State Triangle Law of Vector Addition

There are three laws for vector addition: 

• Triangle law of vectors for the addition of two vectors.

• Parallelogram law of vectors for the addition of two vectors.

• Polygon law of vectors for the addition of more than two vectors.

Vector addition and subtraction are integral parts of mathematical physics. Forces are vectors, and the vector sum of all the individual forces exerted on an object is used to calculate the net force experienced by that object. This is the use of vectors in Newtonian mechanics. Vectors are used in almost every field of physics and the triangular law of vector addition is an important law for their addition. According to the triangular law of vector addition, if two vectors are represented by two sides of a triangle taken in order, then their vector total is represented by the third side of the triangle taken in the opposite direction.

Image: Triangle law of vector addition

The sum, total, or the resultant of any two vectors A and B is represented as,

• $\vec{R}=\vec{A}+\vec{B}$

Derivation of Triangle Law of Vector Addition

Image: Triangle law of vector addition

From the above figure,

P = Vector P

Q = Vector Q

OA = Magnitude of vector P 

AB = Magnitude of vector Q 

R = Sum of vector P and Q using triangle law of vector addition

$\theta=$ The angle between vectors P and Q

Extending the side of OA till point C. So that line BC is perpendicular to OC. The direction of the resultant vector R is given by the angle $\phi$. 

From the right-angled triangle OBC,

$O B^{2}=O C^{2}+B C^{2}$

$O B^{2}=(O A+A C)^{2}+B C^{2}$ …(1)

From the right triangle ABC,

• $\begin{align} \cos \theta &=\dfrac{A C}{A B} ; \sin \theta=\frac{B C}{A B} \\ A C &=A B \cos \theta \\ A C &=Q \cos \theta \\ B C &=A B \sin \theta \\ B C &=Q \sin \theta \\ A C &=Q \cos \theta \quad ; \quad B C=Q \sin \theta\dots(2) \end{align}$

Substituting values from equation(2) in equation (1). We get,

• $\begin{align} R^{2} &=(P+Q \cos \theta)^{2}+(Q \sin \theta)^{2} \\ R^{2} &=P^{2}+Q^{2} \cos ^{2} \theta+2 P Q \cos \theta+Q^{2} \sin ^{2} \theta \\ R^{2} &=P^{2}+2 P Q \cos \theta+Q^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \end{align}$

We already knew that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

• $\begin{align} R^{2} &=P^{2}+2 P Q \cos \theta+Q^{2} \\ R &=\sqrt{P^{2}+2 P Q \cos \theta+Q^{2}} \end{align}$

The magnitude of the resultant vector R is given by the equation,

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

To  find the direction of R, we are taking the right triangle OBC,

• $\tan \phi=\dfrac{B C}{A C}$

From equation (2),

• $\begin{align} \tan \phi &=\dfrac{Q \sin \theta}{(O A+O C)} \\ \tan \phi &=\dfrac{Q \sin \theta}{(P+Q \cos \theta)} \\ \phi &=\tan ^{-1}\left(\dfrac{Q \sin \theta}{(P+Q \cos \theta)}\right) \end{align}$

The direction of the resultant vector of A and B, which in our case is R is given as,

• $\phi=\tan ^{-1}\left(\dfrac{Q \sin \theta}{P+Q \cos \theta}\right)$

Notes on Triangle Law of Vector Addition

• When the magnitude and direction of two vectors can be represented by the two sides of a triangle in the same order, the resultant is represented by the third side of the triangle in the opposite order.

• The magnitude of the resultant vector is given by, 

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

• The direction of the resultant vector R is given by,

• $\phi=\tan ^{-1}\left(\dfrac{Q \sin \theta}{P+Q \cos \theta}\right)$

Where, 

R = Resultant vector

P = Vector P

Q = Vector Q

$\theta=$  The angle between vectors P and Q

$\phi =$ The direction of the resultant vector R

Numerical Examples on Triangle Law of Vector Addition

Example 1: Two vectors having magnitudes of 4 units and 5 units are to be added. The vectors mentioned in this question make an angle of 60° with each other. Find the magnitude and the direction of the resultant vector using the triangle law of vector addition?

Solution: The triangle law of vector addition formula is given as,

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

Let P=4,  and Q=5. 

Given that $\theta=60^{\circ}$, so using this we get,

$\begin{align}&R=\sqrt{4^{2}+5^{2}+2 \times 4 \times 5 \times \cos 60^{\circ}} \\ &\Rightarrow R=\sqrt{16+25+40 \times 0.5} \\ &\Rightarrow R=\sqrt{61} \\ &\Rightarrow R=7.81 \end{align}$

The direction is given as,

• $\begin{align} &\phi=\tan ^{-1}\left(\dfrac{Q \sin \theta}{P+Q \cos \theta}\right) \\ &\Rightarrow \phi=\tan ^{-1}\left(\dfrac{5 \sin 60^{\circ}}{4+5 \cos 60^{\circ}}\right) \\ &\Rightarrow \phi \simeq \tan ^{-1}\left(\dfrac{4.33}{6.5}\right) \\ &\Rightarrow \phi \simeq 33.66^{\circ} \end{align}$

So the resultant of the given vectors is 7.81 units and the direction is approximately 33.66°.

Example 2: The magnitude of the resultant of two vectors having an angle of 60° between them is 8 units. One of the vectors has a magnitude of 2 units and the direction of the resultant is 45°. Find the magnitude of the second vector?

Solution: Let, P=2. 

Given that $\varphi=45^{\circ}$ and $\theta=60^{\circ}$.

We have,

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

Putting the values that we have, we can find the value of Q.

$\begin{align} &\tan 45^{\circ}=\dfrac{0.866 Q}{2+0.5 Q} \\ &\Rightarrow 1 \times(2+0.5 Q)=0.866 Q \\ &\Rightarrow 2=Q(0.866-0.5) \\ &\Rightarrow \dfrac{2}{0.366}=Q \\ &\Rightarrow Q \simeq 5.46 \end{align}$

So, the second vector is approximately equal to 5.46 units.

Conclusion

The triangle law of vector addition is a mathematical concept that is used to find the sum of two vectors. Vector addition and subtraction are integral parts of mathematical physics. A vector is a quantity, or it is also called an object that has both a magnitude and a direction. But a scalar is a quantity that has only magnitude and no direction. The process of adding two or more vectors is known as vector addition. The vectors are added geometrically. Triangle Law, Parallelogram Law and Polygon Law are the three laws for vector addition.

The triangle law for vector addition states that if two vectors are represented by two sides of a triangle taken in order, then their vector sum is represented by the third side of the triangle taken in the opposite direction.