tadiyosyehualashet.files.wordpress.com€¦ · Web viewUnit I Vectors; Lines and Planes 1.1 Vectors 1.1.1 Scalar and Vector Quantities A scalar is a quantity that is determine by

Unit I Vectors; Lines and Planes

Unit I

Vectors; Lines and Planes

1.1 Vectors

1.1.1 Scalar and Vector Quantities

A scalar is a quantity that is determine by its magnitude (its number of units measured in a suitable scale).

Examples 1 Mass, length, temperature, voltage are examples of scalar quantities.

Quantities that have both magnitude and direction are called vectors. A vector is usually represented by

an arrow, the length of the arrow represents the magnitude of the vector and the arrow head indicates the

direction of the vector.

Examples 2 Velocity, acceleration, displacement and force are examples of vector quantities.

When a vector is represented by an arrow, say , the point A is called the initial point (tail) and B is

called the terminal point (head) of the vector. Vectors can also be represented by a single letter (usually

small letter) with a bar over it such as , etc.

Example 3

Notations or - vector A or vector a. - a vector with initial point P and terminal point Q.

- magnitude (or norm) of vector A.

1.1.2 Equality of two Vectors

Definition 1.1 Two non-zero vectors and are said to be equal, denoted = , if

and only if they have the same direction and magnitude, regardless of the

position of their initial points.

Note that: Equality of vectors is transitive.

i.e. For three vectors , and , if = and = , then = .

Definition 1.2 A Vector is called a free vector, provided that its magnitude and direction

are fixed, but its position is indeterminate. If the initial point of a free vector is

fixed, then it is called a localized vector.

Definition 1.3 Two ( free) Vector are equal if and only if they have the same magnitude

and direction.

Definition 1.4 A vector of magnitude ( modulus) unity ( one) is called a unit vector.

a

initial point

1

terminal point


Definition 1.5 Any vector whose magnitude is zero and direction indeterminate is called

a null ( zero) vector. A null vector is denoted by .

Note that: For any non-zero vector A , is a unit vector in the direction of that of vector A.

1.1.3 Vectors in 2 and 3

Definition 1.6 Position Vectors

A non-zero vector in 2 (or 3) is called a position vector if and only if its initial

point is at the origin and its terminal point is anywhere other than the origin.

From this definition, the initial point (tail) of a vector can be anywhere with out changing the direction and

the magnitude of the vector.

Vector Addition

Definition 1.7 Let and be two vectors in a plane. Then the sum + is the

vector represented by .

Triangle Law of Vector Addition

Let and be any two vectors. To find + , join and head to tail. + is the

vector whose initial point is that of and terminal point that of . This law of vector addition

is called triangle law of vector addition.

For any vector , + = .

Theorem 1.1

a) For any two vectors and

+ = + .

b) For any three vectors , and

+ ( + ) = ( + )+ .

For any vector there exists a vector – such that + (– ) = . – , called the opposite of vector ,

has the same magnitude and opposite in direction to that of .

Subtraction of Vectors

For any two vectors and , – is the vector defined by adding and – pictorially

illustrated as follows:

Scalar Multiplication

Definition 1.8 Let be any vector and k be any scalar. k is a vector whose magnitude is

b

a

b

a)( ba

2


times and its direction is that of if k > 0, opposite to that of if

k < 0 and indeterminate if k = 0.

Definition 1.9 For any two vectors and and any two scalars m and n

i) m ( + ) = m + m ii) (m + n) = m + n

iii) (m n) = m (n ) iv) 1 = and 0 =

Components and Coordinate Representation of Vectors.

Definition 1.10 Two vectors and are said to be parallel if = t for some real

number t.

Let and be any two non-zero vectors which are not parallel. Then any vector in the plane of

and can be uniquely expressed as

= s + t

where s and t are scalars.

In this case, we say that is expressed as a linear combination of and . The vectors s and t are

called the component vectors of relative to and respectively and is called a base.

Any pair of non-collinear vectors may be chosen as a base, but the usual and the most convenient choice of

base is a pair of unit position vectors (vectors of unit length) along the positive x-axis and along the positive

y-axis.

Note that: Any position vector is uniquely determined by the coordinates of its terminal point.

Now the position vector (1, 0) is usually denoted by and (0, 1) by . The vectors and , being perpendicular , is called an orthogonal base. Note that: For any non-zero free vector there is a unique position vector such that = .

Now let (x, y) be the terminal point of a position vector . Then can be expressed as:

= x + y .Similarly, if is a free vector with initial point P(x1, y1) and terminal point Q(x2, y2) can be expressed as:

= (x2 – x1) + (y2 – y1) . Notation: = (x, y) represents the position vector with terminal point (x, y).

Note that: = (1, 0) and = (0, 1) are unit position vectors determined by the coordinates of their

terminal points.

Similarly, = (1, 0, 0), = (0, 1, 0) ) and = (0, 0, 1) are mutually perpendicular

unit position vectors in 3.

Now let (x, y, z) be the terminal point of a position vector in 3. Then can be expressed as:

= x + y + z .

The length (norm) of a vector = (x, y, z) is denoted and defined by:

Similarly, if is a free vector with initial point P(x1, y1, z1) and terminal point Q(x2, y2, z2) can be expressed

as:

= (x2 – x1) + (y2 – y1) + (z2 – z1)

Example 4 Let = (2, 0, 5). Find the coordinates of the terminal point of the vector that is

equal to if P (2, 3, 1) is its initial point.

Solution Let Q (x, y, z) be the terminal point of the required vector.

3


Then = (x – 2, y – 3, z – 1) = (2, 0, 5)

x – 2 = 2, y – 3 = 0 and z – 1 = 5

x = 4, y = 3 and z = – 4.

Therefore, (4, 3, – 4) is the terminal point of the required vector.

Example 5 Let = (2, 0, 5) and let P (0, 3, – 6) and Q (– 4, 3, 4) be the initial point and

terminal points of a vector. Find a real number t such that = t

Solution = t (2, 0, 5) = t ( 4, 0, 10)

4 t = 2 and 10 t = 5

t = 0.5.

Therefore, t = 0.5 is the required solution.

The Scalar (Dot) Product

Before we define the dot product of vectors, we need to define what we mean by the angle between two

vectors.

Definition Let and be any two non-zero free vectors, and let (x1, y1) and (x2, y2) be

position vectors associated to and respectively. The angle between and

is defined to be the angle between the two position vectors (x1, y1) and (x2, y2).

Note that:- The angle between any two non-zero position vectors satisfies the condition 0

Further more; if = 0 or = , then the two position vectors are parallel and if = ,

then the two position vectors are perpendicular.

Definition Let and be any two non-zero free vectors. The scalar (dot or inner) product

of and , denoted is defined by:

= Where is the angle between and .

Note that: , and are numbers and hence scalar (dot) product of any two

non-zero vectors is a scalar quantity.

Note that:- i) For any vector , and hence , since = 0

and cos 0 = 1.

ii) The scalar product of any two non-zero perpendicular vectors and is zero.

i.e. = 0, since = and cos = 0.

iii) If is the angle between two non-zero vectors and , then

cos =

Theorem. For any vectors , and , and any scalar k

i) = ii) k ( ) = =

iii) = +

Theorem. If = and = , then

=

Corollary. If = , = are non-zero vectors and is the angle

4


between and , then

Cos =

The Two Important Inequalities.

Let A and B be two vectors.

1. (Cauchy-Schwarz Inequality)

2. + (Triangle Inequality)

Example. Compute the scalar product of

i) = 3 + 4 and = 4 3

i) = 4 + 3 and = 8 6

Solutions. Using the above definition we get:

i) = (3, 4) (4, 3) = 12 12 = 0.

and ii) = ( 4, 3) (8, 6) = 32 18 = 50.

Example. Given: The angle between two unit vectors and is 60. Then find

i) ii) the angle between and +

Solutions. i) = ( + ) ( + )

= + 2 cos 60 +

= 2 (1 + cos 60)

= 3.

Therefore = .

ii) Let be the angle between and + .

Then = + cos 60

= 1 + cos 60

= 1.5. i)

On the other hand = cos

= cos ii)

From i) and ii) we get:

cos = 1.5 cos =

= 30.

Therefore the angle between and + 30.

Example. Given: = and = + . Find the value of k such that

i) + k is orthogonal to

ii) + k is orthogonal to

Solutions. = 1 and = .

i) + k is orthogonal to if and only if ( + k ) = 0.

Now ( + k ) = 0 + k = 0.

1 + k = 0

k = 1.

Therefore k = 1.

ii) + k is orthogonal to if and only if ( + k ) = 0.

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Now ( + k ) = 0 k + = 0.

1 + 2k = 0

k = 0.5.

Therefore k = 0.5.

Example. Find the angle between ( , 1, 1) and the positive x axis.

Solution. Let A = ( , 1, 1) and B = (1, 0, 0) be two position vectors.

Then = cos

Hence cos =

cos =

= .

Therefore the angle between ( , 1, 1) and the positive x axis is .

Direction Angles and Direction Cosines.

Defn Let A = (a1, a2, a3) be a non-zero vector. The angles , and (between 0 and

inclusively) that A makes with the positive x, y and z axes respectively are called

the direction angles of A.

Now take the unit vectors , and . From this definition we get:

cos = , cos = and cos =

Furthermore;

= cos , = cos = and = cos

Therefore A = ( cos + cos + cos )

cos , cos and cos are called the direction cosines of A.

Example. Let A = ( 2, 0, 3). Find the direction cosine of A.

Solution. = = .

Hence cos = , cos = 0 and cos = .

= , = and = .

Therefore , and are the direction cosines of A.

Projection and Resolution of Vectors.

Defn Let A be a non-zero vector. The projection of a vector B onto A, denoted by

is defined as:

6


Note that:- is a vector parallel to A.

Example. Let A = ( 2, 3, 1) and B = (0, 1, 1). Find and .

Solution. = , = and = 2.

Therefore = A and = B.

Theorem

Let A be a non-zero vector. Then for any vector B,

Proof. .

Therefore .

Now let A and B be orthogonal vectors and let C be a vector in the same plane as A and B.

Then we can express C as a linear combination of vectors parallel to A and B as follows:

C = +

In this case, we say that vector C is resolved into vectors parallel to A and B.

Example. Let A = (0, 1, 2), B = (0, 2, 1) and C = (0, 5, 4). Resolve C into vectors parallel

to A and B.

Solution. = 0 and these three vectors lie on the yz plane.

= and = .

Now = 13, = 6, = = .

Hence = and = .

Therefore C = + .

Example. Let A = (1, 0, 3), B = ( 3, 0, 1) and C = (2, 0, 5). Resolve C into vectors parallel

to A and B.

Solution. = 0 and these three vectors lie on the xz plane.

= and = .

Now = 18, = 1, = = .

Hence = and = .

Therefore C = .

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Cross Product

Defn. Let A = (a1, a2, a3) and B = (b1, b2, b3) be two vectors. The cross (Vector) product

of A and B, written A B is defined by:

A B = (a2 b3 b2a3) + (a3 b1 a1 b3) + (a1 b2 a2 b1)

A B is read as “A cross B”.

Now let us see a simple method how to recall the formula for the cross product of A and B

i) The first method.

A B =

ii) The second method.

Example. Let A = (5, 1, 0) and B = (0, 2, 2). Find A B and B A.

Solution. A B =

= 2 + 10 + 10

Therefore A B = 2 + 10 + 10 .

A B =

= 2 10 10

Therefore A B = 2 10 10 .

Remark: = , = and = .

a1 a2 a3

b1 b2 b3

+ + +

a1 a2 a3 a1 a2

b1 b2 b3 b1 b2

5 1 0

0 2 2

0 2 2

5 1 0

8


Properties of Cross Product.

Let A, B, C be vectors and let m be a scalar. Then

i) A B = (B A)

ii) A A =

iii) A (B + C) = (A B) + (A C)

and (A + B) C) = (A C) + (B C)

iv) (m A) B = m (A B) = A (m B).

Remark: Cross Product is not associative.

Example. ( ) = while, ( ) = = .

Theorem. Let A and B be two non-zero vectors.

a) and

Consequently; if , then is orthogonal to both A and B.

b) If is the angle between A and B (0 ), then

= sin .

Proof. i) = a1 (a2 b3 b2a3) + a2 (a3 b1 a1 b3) + a3 (a1 b2 a2 b1)

= a1a2 b3 a1a3b2 + a2a3 b1 a1a2b3 + a1a3b2 a2 a3b1

= (a1a2 b3 a1a2b3) + (a1a3b2 a1a3b2) + (a2a3 b1 + a2 a3b1)

= 0.

= b1 (a2 b3 b2a3) + b2 (a3 b1 a1 b3) + b3 (a1 b2 a2 b1)

= a2 b1b3 a3 b1b2 + a3b1b2 a1b2b3 + a1b2b3 a2b1b3

= (a2 b1b3 a2b1b3) + (a3b1b2 a3 b1b2) + (a1b2b3 a1b2b3)

= 0.

ii) = (a2 b3 b2a3)2 + (a3 b1 a1 b3)2+ (a1 b2 a2 b1)2

= (a22 b3

2 2 a2a3b2b3 + b2

2 a32) + (a1

2 b32 2 a1a3b1b3 + b1

2 a32)

+ (a12 b2

2 2 a1a2b1b2 + a2

2b12)

= a12 (b2

2 + b3

2) + a22 (b1

2+ b32 ) + a3

2 (b12+ b2

2 )

(2 a2a3b2b3 + 2 a1a3b1b3 + 2 a1a2b1b2)

= (a12 +a2

2 + a32) (b1

2 + b22 + b3

2 ) (a1b1 + a2b2 + a3b3)2

= ( cos ) 2

=

= (1 )

=

Therefore = sin .

Corollary. Two non-zero vectors A and B are parallel if and only if = .

Proof. = = 0

9


sin

sin = 0

= 0 or = .

A∥ B.

Therefore Two non-zero vectors A and B are parallel if and only if = .

Example. Let A (3, 2, 2) and B (0, 3, 7).

i) Determine whether A and B are parallel or orthogonal or neither.

ii) Find a vector orthogonal to both A and B.

Solution. i) = (3 0) + (2 3) + ( 2 7) = 2, = and = .

Hence neither 0 nor .

Therefore A and B are neither parallel nor orthogonal.

Remark: is the area of a parallelogram with adjacent sides A and B.

Triple Product.

There are two types of triple products.

i) Scalar triple product.

For any three vectors A, B and C, is called the triple (box or mixed triple)

product of A, B and C.

Example. Show that for any three vectors A, B and C

=

Solution. = cos , where is the angle between A and .

= cos sin , where is the angle between B and

and = cos sin , where is the angle between C and

and is the angle between A and B.

Now cos = sin and sin = cos , because co-functions of complementary angles

are equal.

Therefore = .

ii) Vector Triple Product.

For any three vectors A, B and C, is called the Vector triple product of

A, B and C.

Example. Show that for any three vectors A, B and C

Solution. = sin , where is the angle between A and .

= sin sin , where is the angle between B and

and = sin sin , where is the angle between C and

and is the angle between A and B.

Now sin sin and sin sin .

Therefore .

Remark: For any three vectors A, B and C

, and

are undefined operations.

Some Properties of Triple Products.

For any three vectors A, B and C

i) = =

ii) = “ bac – cab” rule.

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Remark: For any three non-zero vectors A, B and C; is the volume of a

parallelepiped with sides A, B and C.

Lines in .

A line in space is determined by a point p0 (x0, y0, z0) on ℓ and a non-zero vector L parallel to it.

Now let ℓ be a line parallel to a non-zero vector L and let p0 (x0, y0, z0) be a fixed point on ℓ.

Let p (x, y, z) be an arbitrary point on ℓ. We need to express p in terms of p0 and L.

ℓ ∥ L ∥ L

(x x0, y y0, z z0) = t L ; for some t and t 0.

(x, y, z) = (x0, y0, z0) +

= + t L; for some t and t 0;

where = (x, y, z) and = (x0, y0, z0).

Therefore = + t L; for some t and t 0 is the vector form of the equation of a line.

Example. Find a vector equation of the line that contains ( 1, 3, 5) and is parallel to

Solution. Now = and L = .

= ( ) + t ( )

Therefore = is the required vector form of the

equation of the line.

Now let L = be a given non-zero vector and let (x0, y0, z0) be a point on ℓ. Then for any

point (x, y, z) on ℓ that is parallel to L, the vector equation form of ℓ is given by:

= + t L.

Hence (x, y, z) =

x = , y = and z = . (i)

These equations are called the parametric equations of ℓ and t is called the parameter.

Example. Find the parametric equation of the line that contain ( 2, 1, 3) and is parallel to

.Solution. Now (x0, y0, z0) = ( 2, 1, 3) and L = .

Then x = , y = and z = .

Therefore x = , y = and z = is the required solution.

In the above parametric equations of a line ℓ if a, b, and c are non-zero real numbers then

We can express (i) as follows:

; where t .

This form of the equations of a line is called the Symmetric form of the equation of a line.

Example. Find the symmetric equations of the line containing the points P1 (2, 3, 1) and

P2 (5, 0, 4).

Solution. Now take L = = and (x0, y0, z0) = (2, 3, 1).

Therefore ; where t is the required equation.

Example. Find the vector, parametric and symmetric equations of the line containing the point

P ( 3, 4, 5) which is parallel to .

Solution. (x0, y0, z0) = ( 3, 4, 5).

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i) Vector equation

Hence = ( 3, 4, 5) + t (4, 0, 3).

= ( 3 + 4 t, 4, 5 3 t)

= ( 3 + 4 t) + + (5 3 t)


ii) Parametric equations.

(x, y, z) = ( 3 + 4 t, 4, 5 3 t)

x = 3 + 4 t, y = 4 and z = 5 3t ; where t is the required equation.

iii) Symmetric equations.

x = 3 + 4 t, y = 4 and z = 5 3t ; where t

; where t .


Example. Show that the line containing the points (0, 0, 5) and (1, 1, 4) is perpendicular to the

line with equation .

Solution. Let P (0, 0, 5) and Q (1, 1, 4).

We need to show that and = (7, 4, 3) are perpendicular.

Now = 0.

Therefore the two lines are perpendicular.

Distance from a Point to a Line.

Given a line ℓ and a non-zero vector L parallel to ℓ, we wish to determine the distance D between ℓ and a

point P1 (not on ℓ).

To do so choose a point P0 on ℓand let be the angle between ℓ and , where 0 ≤ ≤ π.

Now D = sin , but = sin .

Therefore is the distance of P1 from the line ℓ.

Example. Find the distance D from the point (2, 1, 0) to the line with equation

x = 2, y + 1 = z = t.

Solution. Take any point P0 on the line. Say P0 ( 2, 1, 0).

P1 (2, 1, 0) and L = (0, 1, 1). Hence = (4, 2, 0).

Now = 2 + 4 4 , = 6 and = .

Therefore D = 3 units.

Planes in 3.

Given a point P0 and a non-zero vector , there exists one and only one plane J containing P0 and

perpendicular to .

Let P0 (x0, y0, z0) be a given point, = (a, b, c) be a non-zero vector and let P (x, y, z) be an arbitrary point

on the plane J containing P0 and which is perpendicular to .

is perpendicular to the plane J. is perpendicular to .

(a, b, c) · (x x0, y y0, z z0) = 0.

a (x x0) + b (y y0) + c ( z z0) = 0. (*)

12


Therefore a (x x0) + b (y y0) + c (z z0) = 0 is the equation of the plane containing P0 and perpendicular to

= (a, b, c).

is said to be normal to the plane J. Expanding and rearranging (*) , we get an equivalent equation of the form:

a x + b y + c z = d; where d = · .

Example. Find the equation of the plane that contains the point (5, 1, 2) and has normal to

.

Solution. P0 (5, 1, 2) and = (2, 0, 3).

Now (5, 1, 2) · (2, 0, 3) = 4.

Therefore 2 x 3 z = 4 is the required equation of the plane.

Example. Find the equation of the plane that contains the point (2, 2, 1) and which is

perpendicular to the line with equation has normal to .

Solution. P0 (2, 2, 1) and = (2, 3, ).

Now (2, 2, 1) · (2, 3, ) = .

Therefore 2 x + 3 y + z = is the required equation of the plane.

Note that: i) Three distinct points P0, P1 and P2 in 3 are collinear if and only if

= 0.

ii) Three distinct non-collinear points P0, P1 and P2 in 3 determine a unique plane.

Let P0, P1 and P2 be three distinct non-collinear points in 3. To determine the equation of the plane J that

contains these points, we need to solve:

· ( ) = 0.

Example. Find the equation of the plane that contains (1, 0, 1), (2, 1, 1) and (2, 0, 3).

Solution. Let P0 (1, 0, 1), P1 (2, 1, 1) and P2 (2, 0, 3).

Then = (3, 1, 0) and = (1, 0, 2).

Hence = (2, 6, 1).

Therefore 2 x 6 y + z = 1 is the required equation of the plane.

Distance from a Point to a Plane.

We need to determine the distance D between a point P1 to a plane J whose normal is the non-zero vector

= (a, b, c). To do so choose a point P0 on J and let be the angle between

and , where 0 ≤ ≤ π.

D = cos .

Now = cos .

Hence D = .

Therefore D = for any point P0 on J.

13


Example. Calculate the distance D between the point P1 (2, 3, 1) and the plane

4 x + 2y + z = 0.

Solution. Let P0 (1, 2, 0) be a point on the plane. Now = ( 4, 2, 1) and = (1, 1, 1).

Hence = 3 and = .

Therefore D = units.

Example. Calculate the distance D between the point P1 (2, 3, 1) and the plane that passes

through A( 3, 0, 2), B(1, 1, 2) and C( 1, 1, 1).

Solution. Let P0 ( 3, 0, 2). Now = (5, 3, 3) and = = ( 3, 12, 6).

Hence = 39 and = .

Therefore D = units.

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tadiyosyehualashet.files.wordpress.com€¦ · Web viewUnit I Vectors; Lines and Planes 1.1 Vectors 1.1.1 Scalar and Vector Quantities A scalar is a quantity that is determine by