# Vectors

## Summary

• Unit vector is defined as:  $\widehat { r } \quad =\quad \frac { pi\quad +\quad { qj } }{ \sqrt { { (p) }^{ 2 }\quad +\quad { (q) }^{ 2 } } }$
• Position vector of two points is:  $\bar { AB } \quad =\quad b\quad -\quad a$
• $\left| magnitude \right| \quad =\quad \sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } }$
• Equation of a vector line is  $r\quad =\quad a\quad +\quad \lambda b$  where a is the point on the line and b is the direction
• Angle between two vectors is  $cos\theta \quad =\quad \frac { a.b }{ \left| a \right| \left| b \right| }$

Vector quantities are extremely useful in physics. The important characteristic of a vector quantity is that it has both a magnitude and a direction. Both of these properties must be given in order to specify a vector completely.

#### Scaler & Vector

A scalar quantity is the one which is completely represented by its magnitude, in terms of a given unit i.e mass (kg), energy (joules) etc are examples of a scalar quantity.

Whereas a vector quantity is the one which is completely represented by its magnitude in terms of a given unit and its direction from point of reference i.e acceleration, displacement, velocity etc.

Let r be a given vector. Using the above definition of a vector quantity, we can write vector r as:

$r\quad =\quad \left| r \right| .\widehat { r }$

Where  $\left| r \right|$  is the magnitude of the vector r and  $\widehat { r }$  is the unit vector in the direction of r.

#### Unit vector

A unit vector  $\widehat { r }$  is the vector whose magnitude is 1 unit. It is used to specify the direction of the given vector. Also:

$\widehat { r } \quad =\quad \frac { r }{ \left| r \right| }$   i.e

If a vector is divided by its magnitude (modulus) then we get a unit vector in the direction of that vector.

Unit vectors can be described as i + j, where i is the direction of the x axis and j is the direction of the y axis.

If r = p i +q j, then the unit vector in the direction of r is given by:

$\widehat { r } \quad =\quad \frac { pi\quad +\quad { qj } }{ \sqrt { { (p) }^{ 2 }\quad +\quad { (q) }^{ 2 } } }$

#### Position Vector

If ”O” is the origin then the position of any point A can be determined by the vector $\bar { OA }$  which is called the position vector of A with respect to the origin O.

We can say that if a is the position vector of the point A, then:  $\bar { OA } \quad =\quad a$

Let a and b be two position vectors of the points A and B with respect to the origin O respectively as shown in Fig 1.

Then:

$\bar { OA } \quad =\quad a$  and  $\bar { OB } \quad =\quad b$

Therefore:

$\bar { AB } \quad =\quad \bar { OA } \quad +\quad \bar { OB }$
$=\quad \bar { -OA } \quad +\quad \bar { OB }$
$=\quad -a\quad +\quad b$
$=\quad b\quad -\quad a$

$\bar { AB } \quad =\quad b\quad -\quad a$

Similarly, we can express other vectors in terms of their position vectors i.e:
$\bar { PQ } \quad =\quad q\quad -\quad p$,
$\bar { MN } \quad =\quad n\quad -\quad m$,  etc

#### Magnitude of a vector

For example for a given vector a i + b j, we can find its magnitude using the formula:

$\left| magnitude \right| \quad =\quad \sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } }$

#### Example #1

Q. Find the magnitude of the vector 5 i + 4 j

Solution:

$\left| magnitude \right| \quad =\quad \sqrt { { a }^{ 2 }\quad +\quad { b }^{ 2 } }$

$=\quad \sqrt { { 5 }^{ 2 }\quad +\quad { 4 }^{ 2 } }$

$=\quad \sqrt { 41 } \quad =\quad \left| 6.4 \right|$      Ans

#### Vector Equation of a line

Let’s suppose a line is parallel to a direction vector b and passes through fixed point A with position vector a. Let r be the position vector of a general point on the line.

Thus the equation of the line is defined by $r\quad =\quad a\quad +\quad \lambda b$   where  $\lambda$  is a parameter and corresponds to a point on the line.

$r\quad =\quad a\quad +\quad \lambda b$

#### Example #2

Q. Find the equation of the line through the points (2, -3) in the direction 3 i – 5 j

Solution:

We know the point a = (2,-3) and b = (3, -5)

Substitute value in the above equation we get:

$r\quad =\quad (2i\quad -\quad 3j)\quad +\quad \lambda (3i\quad -\quad 5y)$

#### Scalar Product

When the product of two vectors is scalar, it is called a scalar product or dot product.

Remember these points about the scalar product;

1.      Scalar product is commutative:  $a\quad .\quad b\quad =\quad b\quad .\quad a$

2.      Scalar product is distributive:  $a\quad .\quad (b\quad .\quad c)\quad =\quad a\quad .\quad b\quad +\quad a\quad .\quad c$  where a,  b,  c are three vectors

Suppose we are given two vectors:

$p\quad =\quad { a }_{ 1 }i\quad +\quad { a }_{ 2 }j\quad +\quad { a }_{ 3 }k\quad \quad \quad and\quad \quad \quad q\quad =\quad { b }_{ 1 }i\quad +\quad { b }_{ 2 }j\quad +\quad { b }_{ 3 }k$

Then the scalar product of p and q would be:

$p\quad .\quad q\quad =\quad ({ a }_{ 1 }{ b }_{ 1 })i\quad +\quad ({ a }_{ 2 }{ b }_{ 2 })j\quad +\quad { ({ a }_{ 3 }{ b }_{ 3 }) }k$

#### Angle between two vectors

If  $\theta$  is the angle between two vectors a and b, then:

$a\quad .\quad b\quad =\quad \left| a \right| \quad .\quad \left| b \right| \quad cos\theta$  or can be written as

$cos\theta \quad =\quad \frac { a\quad .\quad b }{ \left| a \right| \left| b \right| }$

Note: When the dot product is equal to zero, it means two vectors are perpendicular to each other as cos 90°=0