Vector Addition

A Vectors is defined as,

"A quantity that has both magnitudes, as well as direction."

For any point P(x, y, z), the vector \overrightarrow{OP} is represented as:

 \overrightarrow{OP}(=\overrightarrow{r}) = x\hat{i} + y \hat{j} + z\hat{k}

Vector addition is a fundamental operation in vector algebra used to find the sum of two or more vectors. It combines the magnitudes and directions of the vectors to produce a single resultant vector.

In vector addition, the vectors are arranged such that:

1. The head of one vector touches the tail of the next vector.
2. A resultant vector is then drawn from the tail of the first vector to the head of the last vector.

The resultant vector represents the combined effect of the original vectors. For example, if two vectors \vec{a} and \vec{b}are added, their sum is expressed as:

\vec{a} + \vec{b}

Conditions for Vector Addition

Various properties of vector addition are,

• We cannot add vectors and scalars together.
• Vectors can be added only if they are of the same nature. For instance, acceleration should be added with only acceleration and not mass.
• Vectors whose resultant have to be calculated behave independently.

Consider two vectors P and Q, where, 

• P = P_xi + P_yj + P_zk
• Q = Q_xi + Q_yj + Q_zk

Then, the formula for the resultant vector is R = P + Q is given by:

R = (P_x + Q_x)i + (P_y + Q_y)j + (P_z + Q_z)k

Vector Addition Calculator

You can use the following calculator to add any two vectors:

Laws of Vector Addition

There are three basic laws of vector addition that are used to add vectors and that include,

• Triangle Law of Vector Addition.
• Parallelogram Law of Vector Addition
• Polygon Law of Vector Addition

Let's understand these laws of vector addition in detail as follows:

Triangle Law of Vector Addition

If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3^rd side of that triangle taken in opposite order.

Derivation of Triangle Law

Consider two vectors P and Q acting on a body and represented both in magnitude and direction by sides OA and AB respectively of a triangle OAB. Let θ be the angle between P and Q. Let R be the resultant of vectors P and Q. Then, according to the triangle law of vector addition, side OB represents the resultant of P and Q.

So, we have R = P + Q.

Now, expand A to C and draw BC perpendicular to OC.

From triangle OCB,

OB²=OC² + BC²

⇒OB² = (OA + AC)² + BC_{2 . . .} (i)

In triangle ACB, 

cos θ = AC/AB

​⇒ AC = AB cos θ = Q cos θ

Also, sin θ = BC/AB

​⇒ BC = AB sin θ = Q sin θ

Magnitude of Resultant Vector

Substituting the value of AC and BC in (i), we get

R² =(P + Q cos θ)² + (Q sin θ)²

⇒R² = P² + 2 PQ cos θ + Q² cos² θ + Q²sin² θ

⇒ R² = P² + 2PQ cos θ + Q²

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

Which is the magnitude of the resultant.

Direction of Resultant Vector

Let Φ be the angle made by the resultant R with P. Then,

From triangle OBC,

tan Φ = BC/OC = BC/(OA + AC)

⇒ tan Φ = Q sin θ/(P + Q cos θ)

\therefore \phi = tan ^ {-1} ( \frac {Q sin\theta} {P+Q cos\theta} )

Which is the direction of the resultant.

Parallelogram Law of Vector Addition

If two vectors acting simultaneously at a point can be represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through that point.

Note: In Euclidean geometry, it is necessary that the parallelogram should have equal opposite sides.

Derivation of Parallelogram Law

Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in the figure.

Let θ be the angle between P and Q and R be the resultant vector. Then, according to the parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.

So, we have R = P + Q.

Now, expand A to C and draw BC perpendicular to OC.

From triangle OCB, 

OB²=OC² + BC²

⇒OB² = (OA + AC)² + BC² _{. . .} (i)

In triangle ABC,

cos θ = AC/AB

​⇒ AC = AB cos θ = OD cos θ = Q cos θ

[∵AB = OD = Q]

Also, sin θ = BC/AB

​⇒ BC = AB sin θ = OD sin θ = Q sin θ

Magnitude of Resultant Vector

Substituting the value of AC and BC in (i), we get

R² = (P + Q cos θ)² + (Q sin θ)²

⇒R² = P² + 2 PQ cos θ + Q² cos² θ + Q²sin² θ

⇒ R² = P² + 2PQ cos θ + Q²

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

Which is the magnitude of the resultant.

Direction of Resultant Vector

Let ø be the angle made by the resultant R with P. Then,

tan Φ = BC/OC = BC/(OA + AC)

⇒ tan Φ = Q sin θ/(P + Q cos θ)

\therefore \phi = tan ^ {-1} ( \frac {Q sin\theta} {P+Q cos\theta} )

Which is the direction of the resultant.

Polygon Law of Vector Addition

Polygon law of vector addition states that,

"Resultant of a number of vectors can be obtained by representing them in magnitude and direction by the sides of a polygon taken in the same order, and then taking the closing side of the polygon in the opposite direction."

In the specific case of vector A, vector B, vector C, and vector D, the results can be obtained by drawing a polygon with the vectors as its sides and then taking the closing side of the polygon in the opposite direction. The magnitude and direction of the resultant will be the same as the magnitude and direction of the closing side of the polygon.

On joining all vectors by connecting one's tail with the other's head, without changing their magnitude and direction we get a Polygon, and the vector joining the tail of the first and the head of the last vector is our Resultant vector (\vec R = \vec A + \vec B + \vec C + \vec D       ).

Vector Addition Formula

The formula for the resultant of addition of two vectors \bold{\vec A}  and \bold{\vec B} is:

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

where,

• \bold{\vec R}   is the Resultant Vector,
• \bold{\vec A}  is the First Vector,
• \bold{\vec B}  is the Second Vector.

The general notation for the addition of vectors is:

If \vec a= <a₁, a₂, a₃> and \vec b= <b₁, b₂, b₃>, then their sum is given as:

\bold{\vec{a}+\vec{b}}= <a₁ + b₁, a₂ + b₂, a₃ + b₃>

Where,

• \vec a is the First Vector,
• \vec b is the First Vector, and 
• <a₁ + b₁, a₂ + b₂, a₃ + b₃> is the Resultant Vector.

Properties of Vector Addition

The vector addition is the sum of multiple (two or more) vectors. Two laws related to the addition of vectors are parallelogram law and triangle law. Similarly, the properties related to vector addition are:

Vector Subtraction

Vector subtraction of two vectors a and b is represented by a - b and it is nothing but adding the negative of vector b to the vector a. i.e., a - b = a + (-b).

Thus, the subtraction of vectors involves the addition of vectors and the negative of a vector. The result of vector subtraction is again a vector.
The following are the rules for subtracting vectors:

• It should be performed between two vectors only (not between one vector and one scalar).
• Both vectors in the subtraction should represent the same physical quantity.

Vector Subtraction Formula

Here are multiple ways of subtracting vectors:

• To subtract two vectors a and b graphically (i.e., to find a - b), just make them coinitial first and then draw a vector from the tip of b to the tip of a.
• We can add -b (the negative of vector b which is obtained by multiplying b with -1) to a to perform the vector subtraction a - b. i.e., a - b = a + (-b).
• If the vectors are in the component form we can just subtract their respective components in the order of subtraction of vectors.

Thus, the addition formula can be applied as:

\bold{\vec{A} – \vec{B} = \sqrt{A^2~+~B^2~-~2AB~cos~\theta}}

Note: -\vec{B} is nothing but \vec B reversed in direction.

Properties of Vector Subtraction

There are various properties of vector subtraction, some of those properties are:

• Any vector subtracted from itself results in a zero vector. i.e., a - a = 0, for any vector a
• The subtraction of vectors is NOT commutative. i.e., a - b is not necessarily equal to b - a
• The vector subtraction is NOT associative. i.e., (a - b) - c does not need to be equal to a - (b - c)
• (a - b) · (a + b) = |a|² - |b|²
• (a - b) · (a - b) = |a - b|² = |a|² + |b|² - 2 a · b

Read More,

• Scalar and Vector
• Dot and Cross Products on Vectors
• Vector Calculus
• Vector Operations
• Triangle Law of Vector Addition

Solved Examples of Vector Addition

Example 1: Find the addition of vectors PQ and QR, where PQ = (3, 4) and QR = (2, 6)

Solution:

We will perform the vector addition by adding their corresponding components

PQ + QR = (3, 4) + (2, 6)
⇒ PQ + QR = (3 + 2, 4 + 6)
⇒ PQ + QR = (5, 10).

Thus, the required answer is (5, 10).

Example 2: If a = <1, -1> and b = <2, 1> then find the unit vector in the direction of addition of vectors a and b.

Solution:

The vector sum is: a + b = <1, -1> + <2, 1> = <1 + 2, -1 + 1> = <3, 0>
Its magnitude is, |a + b| = √(3² + 0²) = √9 = 3.

The unit vector in the direction of vector addition is: (a + b) / |a + b| = <3, 0> / 3 = <1, 0>

Thus, the required unit vector is, <1, 0>.

Example 3: If a = <4, -2, 3> and b = <1, -2, 5> then find a - b.

Solution:

a - b = <4, -2, 3> - <1, -2, 5>
⇒ a - b = <4 - 1, -2 - (-2), 3 - 5>
⇒ a - b = <3, 0, -2>

Therefore, a - b = <3, 0, -2>.

Example 4: Two forces of magnitude 6N and 10N are inclined at an angle of 60° with each other. Calculate the magnitude of the resultant and the angle made by the resultant with 6N force.

Solution:

Let P and Q be two forces with magnitude 6N and 10N respectively and θ be angle between them. Let R be the resultant force.

So, P = 6N, Q = 10N and θ = 60°

We have,

R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta} \\ \Rightarrow R = \sqrt{6^2 + 10^2 + 2.6.10 \cos 60 \degree} \\ \therefore R = \sqrt {196} = 14N

Which is the required magnitude.

Let ø be the angle between P and R. Then,

\tan\phi=\frac{Q\sin\theta}{P+Q\cos\theta} \\ \Rightarrow \tan\phi=\frac{10\sin60\degree}{6+10\cos60\degree} \\ \Rightarrow \tan\phi=\frac{5\sqrt{3}}{11} \\ \therefore \phi=\tan^{-1}({\frac{5\sqrt{3}}{11}})

Which is the required angle.

ayushagnihotri2025

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Article Tags :

• Mathematics
• Class 12
• Maths-Class-12
• Physics-Concepts