Types of Vectors

Types of Vectors: After we have learned ‘Vectors and its Components’ and ‘Magnitude and Direction of a Vector’, we’ll learn the Types of Vectors here. Vectors are of different types according to their magnitude, direction, and position with respect to another given vector, and by the method of representing vectors.

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Following are the different types of vectors:

1. Row Vector

If vector components are written in a horizontal order enclosing in a bracket and separating with space but not comma, the vector represented in such a way is called a row vector. ("OP" ) ⃗ = (x-component y-component) = (x y)

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2. Column Vector

If the components of a vector are written in a vertical order enclosing in a bracket, the vector represented in such a way is called a column vector. (AB) ⃗ = (■(x-component@y-component)) = (■(x_2-x_1@y_2-y_1 )) Hence, the row vector and the column vector are the method of representing a vector in a row or in a column form.

3. Zero vector or Null Vector

A vector having zero magnitude is called a zero or null vector. So, if |(OP) ⃗| = 0, then the vector (OP) ⃗ is called the zero vector. The initial and terminal points of a zero vector coincide i.e. (AA) ⃗ = (BB) ⃗ = ( O ) ⃗. So, a zero vector has no definite direction. Zero vector is generally denoted by ( O ) ⃗. Both the components of a zero vector are zero. So, ( O ) ⃗ = (■(0@0)). Zero vector exists while adding two vectors or subtracting one vector from another.

4. Proper Vector

A vector whose magnitude is not zero is called a proper vector. For example: (AB) ⃗ = (■(3@5)), (PQ) ⃗ = (■(1@0)), etc. are proper vectors.

5. Negative Vector

A vector having the same magnitude as a given vector (OP) ⃗ but direction opposite to it is denoted by - (OP) ⃗ or (PO) ⃗ and is called the negative of (OP) ⃗. The negative vector - (OP) ⃗ means the vector which represents a movement equal in magnitude but opposite in direction to (OP) ⃗. It means if (OP) ⃗ displace the point O to the point P, then - (OP) ⃗ displaces the point P to the point O with magnitude equal to the magnitude of (OP) ⃗. If (OP) ⃗ = (■(x@y)), then - (OP) ⃗ = (■(-x@-y)).

6. Equal Vectors

Two vectors are said to be equal if they have same magnitude and same direction. Let two vectors (AB) ⃗ and (CD) ⃗ are in same direction and |(AB) ⃗| = |(CD) ⃗|. Then (AB) ⃗ and (CD) ⃗ are equal and we write, (AB) ⃗ = (CD) ⃗. If (AB) ⃗ = (■(a@b)), (CD) ⃗ = (■(c@d)) and (AB) ⃗ = (CD) ⃗, then a = c and b = d.

7. Unequal Vectors

If two vectors are not equal, then they are said to be unequal vectors. Some vectors may have same direction but different magnitude. Some vectors may have same magnitude but different direction and some vectors may have different magnitude and different direction. All these vectors are unequal vectors.

8. Unit Vector

A unit vector is a vector whose magnitude is unity. So, if |(OP) ⃗| = 1, then (OP) ⃗ is said to be a unit vector. Some examples of unit vector are (■(0@1)), (■(1@0)), (■(1/√2@1/√2)), etc. Every non-zero vector has a unit vector along its own direction. So, unit vector is not fixed. Each unit vector has the same magnitude i.e. unity but their direction may vary. The unit vector along the direction of vector (OP) ⃗ is denoted by (OP) ̂ and it is defined by (OP) ̂ = (OP) ⃗/(|(OP) ⃗|). Similarly, the unit vector along the direction of vector ( a ) ⃗ is denoted by a ̂ and is defined by a ̂ = ( a ) ⃗/(|( a ) ⃗|). So, if (OP) ⃗ = (■(x@y)), then |(OP) ⃗| = √(x^2+y^2 ) and (OP) ̂ = (OP) ⃗/(|(OP) ⃗|) = 1/√(x^2+y^2 ) (■(x@y)) = (■(x/√(x^2+y^2 )@y/√(x^2+y^2 ))). We read a ̂ as ‘a cap’. There are two special types of unit vectors in two dimensional vector system. The unit vector along X-axis is denoted by ( "i" ) ⃗ and the unit vector along Y-axis is denoted by ( "j" ) ⃗ where ( "i" ) ⃗ = (■(1@0)) and ( "j" ) ⃗ = (■(0@1)).

9. Parallel or Collinear Vectors

Two vectors are said to be parallel or collinear if they are of same or opposite direction whatever may be their magnitudes. That is, two vectors ( a ) ⃗ and ( b ) ⃗ are said to be parallel if ( a ) ⃗ = k( b ) ⃗, where k is any positive or negative number.

10. Like Vectors

Two vectors are said to be like vectors if they have same directions whatever may be their magnitudes. i.e. Two vectors ( a ) ⃗ and ( b ) ⃗ are said to be like if ( a ) ⃗ = k( b ) ⃗ for k > 0.

11. Unlike Vectors

Two vectors are said to be unlike vectors if they have opposite directions whatever may be their magnitudes. i.e. Two vectors ( u ) ⃗ and ( v ) ⃗ are said to be unlike if ( u ) ⃗ = k( v ) ⃗ for k < 0.

12. Perpendicular or Orthogonal Vectors

Two vectors are said to be perpendicular or orthogonal vectors if angle between them is 90°. The vectors: (AB) ⃗ = (■(2@4)) and (PQ) ⃗ = (■(-2@1)) are orthogonal as, direction of (AB) ⃗ = tan-1(4/2) = 64° (appr.) direction of (PQ) ⃗ = tan-1(1/(-2)) = 154° (appr.) And, angle between (AB) ⃗ and (PQ) ⃗ = 154° – 64° = 90°.

13. Localised Vector

A vector which passes through a given point and which is parallel to the given vector is said to be a localised vector. Let ( a ) ⃗ be the given vector and ( b ) ⃗ be the vector drawn parallel to ( a ) ⃗ from the given point A in a plane. Then, ( b ) ⃗ is called the localised vector.

14. Co-initial Vectors

The vectors which have same initial point are called Co-initial vectors. (AB) ⃗ and (AC) ⃗ are Co-initial vectors. They have same initial point A.

15. Coplanar Vectors

Two or more than two vectors are said to be the coplanar if they lie on a same plane or are parallel to the same plane, otherwise they are said to be non-coplanar vectors.

Worked Out Examples

Example 1: If (AB) ⃗ displaces the point A(2, 5) to B(6, 1) and (PQ) ⃗ displaces the point P(4, -1) to Q(8, -5), (a) Find the components of (AB) ⃗ and (PQ) ⃗. (b) Express (AB) ⃗ and (PQ) ⃗ as column vectors. Solution: (a) x-component of (AB) ⃗ = 6 – 2 = 4 y-component of (AB) ⃗ = 1 – 5 = -4 and, x-component of (PQ) ⃗ = 8 – 4 = 4 y-component of (PQ) ⃗ = -5 + 1 = -4 (b) Hence, (AB) ⃗ = (■(4@-4)) and (PQ) ⃗ = (■(4@-4))

Example 2: If vector (PQ) ⃗ displaces the point P(2, -1) to Q(5, 7), find a unit vector along (PQ) ⃗. Solution: As vector (PQ) ⃗ displaces the point P(2, -1) to Q(5, 7), (PQ) ⃗ = (■(5-2@7+1)) = (■(3@8)) The magnitude of (PQ) ⃗ = |(PQ) ⃗| = √(3^2+8^2 ) = √(9+64) = √73 units. Hence, (PQ) ⃗ is not a unit vector. The unit vector along (PQ) ⃗ = ((PQ) ⃗ )/(|(PQ) ⃗ |) = 1/√73 (■(3@8)) = (■(3/√73@8/√73))

Example 3: If P, Q, R and S are four points with co-ordinates (2, -2), (6, 4), (x, y) and (3, -5) respectively such that (PQ) ⃗ = (RS) ⃗, find the co-ordinates of R. Solution: The co-ordinates of P and Q are (2, -2) and (6, 4) respectively. Hence, (PQ) ⃗ = (■(6-2@4+2)) = (■(4@6)) The co-ordinates of R and S are (x, y) and (3, -5) respectively. Hence, (RS) ⃗ = (■(3-x@-5-y)) We have, (PQ) ⃗ = (RS) ⃗ i.e. (■(4@6)) = (■(3-x@-5-y)) i.e. 4 = 3 – x ⟹ x = -1, and 6 = -5 – y ⟹ y = -11 ∴ x = -1 and y = -11.

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