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A vector is a quantity that has a magnitude and a direction. Both these properties must be given in order to specify a vector completely. Different Laws of Vector addition have been included in this article. An example of a vector is displacement, which is the distance traveled from one point to the other in a particular direction. Addition of Vectors can be done by following the Triangle Law and Parallelogram Law.
Read Also: NCERT Solutions For Class 12 Mathematics Chapter 10 Vector Algebra
| Table of Content |
Key Terms: Triangle Law of Vector Addition, Vectors, Magnitude, diagonal, addition of vectors, parallelogram, triangle.
Triangle Law of Vector Addition
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When two vectors positioned at two adjacent sides of a triangle, the sum of the two Vectors in the same order, will be represented by the third side of the triangle taken in the reverse order.
If A→and B→ are two vectors in the same direction, then A→ + B→ is the sum of vector A→ and B→.
Triangle Law of Vector Addition
Read Also: magnitude of the vector
The video below explains this:
Addition of Vectors Detailed explanation:
Parallelogram Law of Vector Addition
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If two vectors are represented as the adjacent side of a parallelogram, then the sum of the vectors will be represented as the diagonal of the parallelogram passing through the common point.
If A→ and B→ are the two adjacent vectors of the parallelogram, then the diagonal passing through the common point will be A→ +B→.
Parallelogram Law of Vector Addition
Properties of Vector Addition
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Properties of Vector Addition have been explained below:
Commutative Property
The commutative property of addition vector states that “A→ + B→ = B→ + A→”
Associative Property
The associative property of addition vector states that,
| (A→ + B→) + C→ = A→ + (B→ + C→) |
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An additive Identity is a vector value when summed gives an identical result.Additive Identity
A→ + 0 = A→; here 0 is Additive Identity of A vector.
Additive Inverse
An additive Inverse is a vector value when summed gives zero.
A→ + (-A→) = 0; here (-A→) is Additive Inverse of A vector.
Read More: Types of Vector
Multiplication of Vector by a Scalar
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While multiplying a vector by a scalar, multiple each component by a scalar.
If P = (P1, P2) has magnitude of |P| and direction d,
| ηP=η (P1, P2) = (ηP1, ηP2) |
|ηP| is the magnitude
d is the direction
If η is negative, then the direction of ηP is opposite of d.
Read More: multiplication of a vector by a scalar
Component Form of Vector
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If there are three axis, X, Y, and Z. One unit in the direction of X-axis, Y-axis and Z-axis are termed as î, \(\hat{j}\), and \(\hat{k}\) respectively.
Component Form of Vector
Now to understand the component form of a vector, look at the below diagram.
Understanding Component Form with Diagram
There is X, Y, Z vector, and P is a component point with O as an origin. Then O to P is OP vector.
Now the vector component is for OX is î, OY is \(\hat{j}\) and OZ is\(\hat{k}\).
So the component form of OP= X î + Y \(\hat{j}\) + Z\(\hat{k}\)
Read More: vector product of two vectors
Unit Vector
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A vector with a magnitude of one is termed a Unit vector. A unit vector is denoted by ‘^’, called a cap or hat.
â = a / |a|
Where |a| is a magnitude of vector a
Unit vectors usually form the base of vector space. Vectors in the space can be expressed by linear combination.
To change a vector in a unit vector we divide the vector by its magnitude. Let’s take XYZ coordinates.
A= xi+ yj+zk
The formula for the magnitude of a vector is |a→|=\(\sqrt {x^2+y^2+z^2}\)
The formula for Unit vector is Unit vector = vector/vector’s magnitude
Read More: Angle Between Two Vectors
Conditions of Collinearity
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Vectors parallel to one line or are drawn on one line are termed as collinear vectors.
Conditions of Collinearity
There are three conditions of collinearity.
- Two vectors X→ and Y→are linear if there exists a number n
X→= η. y→
- Two vectors are collinear if the relation of their coordinates is equal. Not valid if one of the components of the vector is 0.
- Two vectors are collinear if their cross products are equal to zero vectors. Applicable for three-dimensional problems.
Read More: Area of Segment of a Circle
Vector Joining Two Points
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Let A1(X1, Y1, Z1) and A2(X2, Y2, Z2) be the initial point and terminal point respectively, then the vector joining A1 and A2 is given as vector A1 A2 →.
Joining the points A1 and A2 with the origin O, and applying triangle law of addition.
Then from triangle OA1 A2 (figure): OA1 → + A1A2 → = OA2 →
The vector will be always specified as- (Terminal point – Initial point).
This implies A1A2 → = OA2 → - OA1 →
Therefore A1A2 →= ((X2 î + Y2 \(\hat {j}\) + Z2 \(\hat {k}\)) − (X1 î + Y1 \(\hat {j}\) + Z1 \(\hat {k}\)))
= ( X2 - X1) î + (Y2 – Y1) \(\hat {j}\) +( Z2 - Z1) \(\hat {k}\)
The magnitude of A1A2→ is given as:-
I A1A2 → I= S ( (X2 - X1)2+ (Y2 – Y1)2+( Z2 - Z1) 2)
Vector Joining Two Points
Read Also: Centroid of a Triangle
Things to Remember
- A vector has a magnitude and direction.
- Vectors are added geometrically.
- Commutative law states that order of addition is not specific; A+B = B+A.
- According to associative law, the sum of three vectors does not rely on which pair of vectors is first added.
- Two vectors can be summed only if they belong to the same unit.
- A vector with a magnitude of one is termed a Unit vector.
- An additive Identity is a vector value when summed gives an identical result.
- An additive Inverse is a vector value when summed gives zero.
- Two vectors X→ and Y→are linear if there exists a number n X→= η. y→
- Two vectors are collinear if the relation of their coordinates is equal. Not valid if one of the components of the vector is 0.
- Two vectors are collinear if their cross products are equal to zero vectors. Applicable for three-dimensional problems.
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Sample Questions
Ques. Find the vector with the initial point (4,2) and terminal point (-10, 14). (1 Mark)
Ans. AB→= (-10-4) î + (14-2) \(\hat {j}\)
AB→= -6 î + 12 \(\hat {j}\)
Ques. Find the addition of vectors AB and BC, where AB = 6,8 and BC = 7, 2. (1 Mark)
Ans. AB+ BC= (6,8) + (7,2)
AB+ BC= (6+7, 8+2)
AB+ BC= (13, 10)
Ques. PQRS is a quadrilateral, simplify the following. (3 Marks)
Ans. 
SR→ + RP→ =
QS→+SR→+RP→ =
SR→ + RP→ = SP→ (Triangle Law of Vector Addition)
QS→+SR→+RP→ = (QS→+SR→)+RP→ (Associative Law)
=(QR→)+RP→ (Triangle Law of Vector Addition)
Triangle law of vector addition = QP→
Ques. Find the value of N at which the vector A→ = (6,4) and B→ = (18,N) are collinear. (3 Marks)
Ans. Px/Qx =Py/Py
6/18 = 4/N
N= 4X18/6
N= 12
Vector A→ and B→ are collinear when N=12
Ques. Which of the vectors P=(1,4); Q=(4,16); R=(5,9) are collinear. (3 Marks)
Ans. Px/Qx =Py/Py
Vector P→ and Q→ are collinear because 1/4 = 4/16
Vector P→ and R→ are not collinear because 1/4 not equal to 5/9
Vector Q→ and R→ are not collinear because 4/16 is not equal to 5/9
Ques. Find the value of n and m at which the vectors P→= (6,4,m) and Q→= (18,n,24) are collinear. (3 Marks)
Ans. Ax/Bx = Ay/By = Az/Bz
6/18 = 4/n = m/24
6/18 = 4/n
6/18= m/24
n = 18X4/6 = 12
m = 6X24/18 = 8
Vectors Q→ and P→ are collinear when n = 12 and m = 8
Ques. Find the direction cosines of the vector joining the points A(2,4,-6) and B(-2,-4,2). (5 Marks)
Ans. The given points are A (2,4,-6) and B(-2,-4,2)
Therefore AB→ = (-2-2) î + (-4-4) \(\hat {j}\) + (2- (-6)) \(\hat {k}\)
= (-4) î +(-8) \(\hat {j}\) +(2+6) \(\hat {k}\)
=(-4) î +(-8) \(\hat {j}\) +(8) \(\hat {k}\)
= -4 î -8 \(\hat {j}\) +8 \(\hat {k}\)
Therefore |AB|= √ (-4)2 + (-8) 2 +(8) 2
= √(16)+(64)+(64)
=√144
=12
Hence direction of cosines of AB→ are (-4/12), (-8/12), (8/12)
=(-1/3), (-2/3), (-2/3)
= -1/3, -2/3, 2/3
Ques. Two vectors are given along with either component; E=(2,3) and F = (2,-2). Calculate the magnitude and angle of the sum G using the components. (5 Marks)
Ans. In the E→, Ex =2 and Ey=3
In the F→ Fx=2 and Fy =-2
Add the two vectors
E+F = (2,3) + (2, -2) = (4, 1)
It can be written as G= (4,1)
Here G→, Gx =4 and Gy=1
The Magnitude of the resultant vector can be calculated as
|G→| = √((Gx)2+(Gy)2)
|G→| = √((4)2+(1)2)
= √(16+1)
= √(17)
=4.123 units
The angel is calculated as follows,
Φ= tan -1 (Gy/Gx)
Φ=tan-1 (1/4)
Φ=14.04 degrees
Thus the magnitude of the resultant vector is, |G→| = 4.123 units
And the angel Φ=14.04 degrees.
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