Calculus 2 Chapter10 vectors in space

3/15/2015

1

VECTORS AND THE GEOMETRY

OF SPACE 10.1 &10.2: VECTORS IN THE PLANE (Two dimensional)

AND IN SPACE (Three dimensional)

We denote the directed line segment extending from the

point P (the initial point) to the point Q (the terminal point)

by

We refer to the length of as its magnitude, denoted

We use the term vector to describe any quantity that has

both a magnitude and a direction.

Prepared by Dr. F.G.A

Sharjah University

3/15/2015

2

10.1 &10.2: VECTORS IN THE PLANE AND IN SPACE

Three Dimensional Space: A point in three-dimensional Euclidean space, , is specified as an ordered triple (a, b, c), where the

coordinates a, b and c represent the

distance from the origin along each of

three coordinates axes (x, y and z).

Ex 1: Plot the point (3, 2, 4). Sol:


Sharjah University


OF SPACE


Ex 2: Plot the points (2, 3, 2) and (-1,-1,-3).

Remark: In the coordinate axes only x and y, is specified

as an ordered of (a, b), where the coordinates a and b

represent the distance from the origin along each of two

coordinates axes (x and y).

Scalar Multiplication: If we multiply a vector by a scalar

c > 0, the resulting vector will

have the same direction as ,

but will have magnitude


Sharjah University

2

u

u


OF SPACE

3/15/2015

3



Sharjah University

On the other hand, multiplying a vector by a

scalar c < 0 will result in a vector with opposite

direction from and magnitude

A vector with its initial point located at the origin is

called a position vector.

u

u


OF SPACE


The position vector with initial point at the

origin (0,0,0) and terminal point A at the

point is denoted by a1, a2 & a3 are the

components of .

a1 first component ;

a2 second component

a3 Third component.

The magnitude of the position vector a may be written as

Remark: In we have two components a1 and a2. The

magnitude written as

The following table provides us with some important

vectors and information: Prepared by Dr. F.G.A

Sharjah University

a

a

2

2 2 2

1 2 3a a a a

2 2

1 2a a a


OF SPACE

3/15/2015

4



Sharjah University

VECTORS IN THE PLANE VECTORS IN THE IN SPACE

Two position vectors, = a1, a2 and

= b1, b2, are equal, i.e., , if

and only if their components are

equal, i.e., if a1 = b1 and a2 = b2.

Two position vectors, = a1, a2, a3

and = b1, b2, b3, are equal, i.e.,

, if and only if their components

are equal. i.e., a1 = b1, a2 = b2 & a3 = b3.

The zero vector is defined to be

; it is the only vector with zero

length.

The zero vector is defined to be

; it is the only vector with

zero length.

We define the additive inverse of

a vector to be . This

says that the vector is a vector

with the opposite direction as and

same length

We define the additive inverse of

a vector to be This

says that the vector is a vector

with the opposite direction as and

same length

a

b a b

a

ba b

0 0,0 0 0,0,0

3

a

a

1 2,a a a

a

a

a

a

1 2 3, ,a a a a

a

a

1 2 3

1 2 3

, ,

1 , , 1

a a a a

a a a a a

1 2

1 2

,

1 , 1

a a a

a a a a


OF SPACE


Two vectors having the same or opposite direction are called

parallel. The vectors and are parallel if and only if ,

for some scalar c. Note: The zero vector is considered parallel to every vector.

EX: Determine whether the given pair of vectors is parallel:

(a) = 2, 3 and = 4, 5,

(b) = 2, 3 and = 4,6. Sol:

(a) = 2, 3 and = 4, 5, If , then

That is, 4 = 2c (so that c = 2) and 5 = 3c (so that c = 5/3).

This is a contradiction, thus, and are not

parallel. Prepared by Dr. F.G.A

Sharjah University

a b b ca

a

a

b

b

a b b ca

a b


OF SPACE

3/15/2015

5


(b) = 2, 3 and = 4,6

If , then

That is, we have 4 = 2c (so that c = 2) and 6 = 3c (which again leads us to c = 2). This says that 2 = 4,6 = and so, 2, 3 and 4,6 are parallel.

Two-Dimensional Position Vectors: We denote the set of

all position vectors in two-dimensional space by


Sharjah University

a b

b ca

a b


OF SPACE


Vectors in Space: The position vector with terminal point

at A(a1, a2, a3) and initial point at the origin (0,0,0) is denoted

by a1, a2, a3.

We denote the set of all three-dimensional position vectors

by

The vector with initial point at P(a1, a2, a3) and terminal point

at Q(b1, b2, b3) corresponds to the position vector

For any two points A(x1, y1) and B(x2, y2), the vector

corresponds to the position vector x2 x1, y2 y1.


Sharjah University


OF SPACE

a

3/15/2015

6


Ex: Find the vector with initial point at A(2,3) and terminal

point B(3,-1)

Sol:

Ex: Find the vector with initial point at A(2,3,1) and terminal

point B(3,-1,-1)

The distance formula for a point in is given by

The distance formula for a point in is given by


Sharjah University

3 2, 1 3 1, 4AB

2 2 2

1 1 1 2 2 2 2 1 2 1 2 1, , , , ,d x y z x y z x x y y z z

3

2

2 2

1 1 2 2 2 1 2 1, , ,d x y x y x x y y


OF SPACE


EX: Find the distance between the points (1, 3, 5) and (5, 2, 3). SOL: From the distance formula, we have

Exercises:-

Find the distance between the given points

1- (0, 3, 1), (3, 6, 8) 2- (5,4,7),(2.-3.6)

3-(-1,-1,-1),(2,2,1) Prepared by Dr. F.G.A

Sharjah University


OF SPACE

3/15/2015

7


Operations on Position Vectors:- For vectors and , and

any scalar c we have


Sharjah University

VECTORS IN THE PLANE VECTORS IN THE IN SPACE

addition:

addition:

subtraction:

subtraction:

scalar multiplication:

Further, we have

scalar multiplication:

Further, we have

a b

1 2 1 2

1 1 2 2

, ,

,

a b a a b b

a b a b

1 2 3 1 2 3

1 1 2 2 3 3

, , , ,

, ,

a b a a a b b b

a b a b a b

1 2 1 2

1 1 2 2

, ,

,

a b a a b b

a b a b

1 2 3 1 2 3

1 1 2 2 3 3

, , , ,

, ,

a b a a a b b b

a b a b a b

1 2 1 2, ,ca c a a ca ca 1 2 3 1 2 3, , , ,ca c a a a ca ca ca

ca c a ca c a


OF SPACE


EX: Let and vectors in .

Find:-

1- if

Sol:

2-

Sol:


Sharjah University

2, 1,5 , 4,3,1u v 6,2,0w 3

2 3x u v w x

2 2 2, 1,5 4, 2,10

3 4,3,1 3 6,2,0 4,3,1 18,6,0 14,9,1

2 3 4, 2,10 14,9,1 18, 11,9

u

v w

x u v w

3x w

3 3 18, 11,9 6,2,0

54, 33,27 6,2,0 48, 31,27

x w


OF SPACE

3/15/2015

8


3-

Sol:

Exercises:-

Compute


Sharjah University

2u v x

2 4, 2,10 4,3,1 48, 31,27 48, 36,36u v x


OF SPACE

1. 2 5

2. 3 3

3. 10 4

If 2 4 + , 4 + + 4

a b

a b

a b

a i j k b i j k


Theorem: For any vectors and in V3, and any scalars d

in the following hold:


Sharjah University

,a b c

1.

2.

3. 0

4. 0

5.

6. 1 & 0 0 1& 0

a b b a Commutativaty

a b c a b c Associativity

a a Zero Vector

a a Additive Invers

d a b da db Distributive Law

a a a Multiplication by


OF SPACE

3/15/2015

9


Standard Basis Vectors: We define the standard basis vectors and by

and form a basis for V3, since we can write any vector

uniquely in terms of and , as follows:


Sharjah University

,i j k

1,0,0 , 0,1,0 & 0,0,1i j k

,i j k

3a V ,i j k

1 2 3 1 2 3, ,a a a a a i a j a k


OF SPACE


Unit Vectors:

For any nonzero position vector , a unit vector

having the same direction as is given by

The basis vectors are unit vectors, since

EX: Find a unit vector in the same direction as 1,2, 3 and write 1,2, 3 as the product of its magnitude and a unit vector.

Sol: First, we find the magnitude of the vector: Prepared by Dr. F.G.A

Sharjah University

1 2 3, ,a a a a

a

1u a

a

1i j k


OF SPACE

3/15/2015

10


The unit vector is

Further,

Exercises:-

Find two unit vectors parallel to the vectors


Sharjah University

1 1 1 2 31, 2,3 , , .

14 14 14 14u a

a


OF SPACE

1,8, 3 , 12,2, 12 9,3, 5and


Finding The Equation of a Sphere: A sphere is the set of all points

whose distance from a fixed

point (the center) is constant;

that is here, all points (x, y, z)

whose distance from (a, b, c) is r.

We have

Squaring both sides gives us

the standard form of the equation of a sphere. Prepared by Dr. F.G.A

Sharjah University

2 2 2

, , , , ,d x y z a b c

x a y b z c r


OF SPACE

3/15/2015

11


EX: Find the geometric shape described by the equation:

Sol:

Completing the squares in each variable, we have

Adding 9 to both sides gives us a sphere

which has a radius of 3 and center (2, 4, 5). Prepared by Dr. F.G.A

Sharjah University


OF SPACE

10.3: THE DOT PRODUCT

Definition: The dot product of two vectors and

in V3 is defined by

Likewise, the dot product of two vectors in V2 is defined by

Remark: The dot product of two vectors is a scalar (i.e., a

number, not a vector). For this reason, the dot product is also

called the scalar product.


Sharjah University

1 2 3, ,a a a a

1 2 3, ,b b b b

1 2 3 1 2 3 1 1 2 2 3 3, , , , .a b a a a b b b a b a b a b

1 2 1 2 1 1 2 2, , .a b a a b b a b a b


OF SPACE

3/15/2015

12


Ex: Compute the dot product for and

.

Sol:

Ex: Find the dot product of the two vectors and

.

Sol:

Exercises:- Compute for


Sharjah University

a b 1,2,3a

5, 3,4b

1,2,3 5, 3,4 1 5 2 3 3 4 11a b

2 5a i j

3 6b i j

2, 5 3,6 2 3 5 6 6 30 24a b


OF SPACE

a b

1. 1,2,3 , 5, 3,4

2. 8 8 9 , 2 6 7

a b

a i j k b i j k


Theorem: For any vectors and in V3, and any scalars d

in the following hold:


Sharjah University

2

1.

2.

3.

4. 0 0

5.

6.

7.

a b b a Commutativaty

a b c a b a c Distributive Law

da b d a b a db

a

a a a

a b a b Cauchy Schwartz Inequality

a b a b Triangle Inequality

,a b c


OF SPACE

3/15/2015

13


Angle Between Vectors in

For two nonzero vectors and in V3, we define the angle (0 ) between the vectors to be the smaller angle between and , formed by placing their initial points at the

same point.

If and have the same direction,

then = 0; if and have opposite directions, then = .

We say that and are orthogonal

(or perpendicular) if = /2 .


Sharjah University

a b

a ba b

a b


OF SPACE

a b


Theorem: Let be the angle between nonzero vectors and . Then

The angle between two vectors and given by

Ex: Find the angle between the vectors = 2, 1,3 and = 1, 5, 6 .

Sol:


Sharjah University

cosa b a b

a

b

a b

1cosa b

a b

a

b

1cos cosa b

a b a ba b


OF SPACE

3/15/2015

14


Ex: Let and . Find

1- 2- 3- Find the angle between and .


Sharjah University

2 2 2

2 2 2

1 1

2,1, 3 1,5,6 2 1 1 5 3 6 2 5 18 11

2,1, 3 2 1 3 4 1 9 14

1,5,6 1 5 6 1 25 36 62

11cos cos 1.953 112

14 62

a b

a

b

a bradians

a b

3 2 2a i j k 5 2b i j k

a b a b a b


OF SPACE


Ex: Find the angle between and

Theorem:

Two vectors and are orthogonal (perpendicular) if and

only if

EX: Determine whether the following pairs of vectors are

orthogonal:

1- = 1, 3,5 and = 2, 3, 10 Sol:

and are not orthogonal.


Sharjah University

6 3 2v i j k 2 2u i j k

a b

0.a b

a b

1,3, 5 2,3,10 2 9 50 39 0a b

a b


OF SPACE

3/15/2015

15


2- = 4, 2,1 and = 2, 3, 14. Sol:

and are orthogonal.

EX: Determine whether and are

orthogonal or not.

Sol:

and are orthogonal.

Remark: is orthogonal to any vector , since


Sharjah University

a b

4,2, 1 2,3,14 8 6 14 0a b

a b

2 4v j k 3 2u i j k

3, 2,1 0,2,4 3 0 2 2 1 4 0 4 4 0u v u v

a0

1 2 30 0,0,0 , , 0a a a a


OF SPACE


Components:

Let be the angle between two nonzero position vectors and .

Drop a perpendicular line segment

from the terminal point of to the

line containing the vector , then

the base of the triangle has length given

by .

The component of along is

written as


Sharjah University

a

b

a

b

cosa

Component of aalong ba b

cosb

Comp a a


OF SPACE

a

bcosa

3/15/2015

16


But Projections:

If the vector represents a force, we are

often interested in finding a force vector

parallel to having the same component

along as .


Sharjah University

cosb

Comp a a a a b

a.

a b

bb

cosa b

a b

Projection of onto ba

b a

b

a


OF SPACE

b

a

Projba


We call this vector the projection of onto , denoted

where represents a unit vector in the direction of .

Ex: For = 2, 3 and = 1, 5, find the component of along and the projection of onto .


Sharjah University

Projbaba

2

Proj

Proj

b b

b

ba Comp a

b

a b b a ba b

b b b

b

bb

b

aba

ba


OF SPACE

3/15/2015

17


Sol:


Sharjah University

2 2

2,3 1,5 2 15 13

1,5 261 5b

a bComp a

b

1,513

Proj26 26

13 1 1 51,5 1,5 ,

26 2 2 2

b b

ba Comp a

b


OF SPACE


Ex: For and , find the component

of along and the projection of onto .

Sol:


Sharjah University

v

2 2b i j k 6 3 2u i j k

u vu

2 2 2

6,3,2 1, 2, 2

1, 2, 2

6 1 3 2 2 2 6 6 4 4

9 91 2 2

v

u vComp u

v

1, 2, 24

Proj9 9

4 4 8 8 4 8 81, 2, 2 , ,

9 9 9 9 9 9 9

v v

vu Comp u

v

i j k


OF SPACE

3/15/2015

18


Work:

Definition: The work done by the force as its point of application move along the vector is defined by

Ex: Find the work done by the force and the

direction of the vector

Sol:


Sharjah University

PQ

PR

w PQ PR

P

Q

R Direction of Motion

5 2F i j

3v i j

5 2 3 11w F V i j i j


OF SPACE


Ex: If the handle makes an angle of /4 with the horizontal and you pull the wagon

along a flat surface for 1 mile (5280 feet),

find the work done.

Exercises:-

Find the for the following vectors:


Sharjah University


OF SPACE

nd Projb b

Comp a a a

1. 3, 9 , 1,7

2. 3 9 6 , 5 4 7

3. 2 5 6 , 9 5 2

a b

a i j k b i j k

a i j k b i j k

3/15/2015

19

10.4: THE CROSS PRODUCT

Definition: The determinant of a 2 2 matrix of real numbers is defined by

EX: Find the determinate of

Sol:


Sharjah University

2 1

4 3

2 1

2 3 4 1 6 4 104 3


OF SPACE


Definition: The determinant of a 3 3 matrix of real numbers is defined as follows:


Sharjah University


OF SPACE

3/15/2015

20


Ex:

Sol:


Sharjah University


OF SPACE


Ex: Evaluate the determinate

Definition: For two vectors and in

, we define the cross product (or vector product) of

and to be


Sharjah University

5 3 1

2 1 1

4 3 1

3V

1 2 3, ,a a a a 1 2 3, ,b b b b

a

b

2 3 1 3 1 2

1 2 3

2 3 1 3 1 2

1 2 3

i j ka a a a a a

a b a a a i j kb b b b b b

b b b


OF SPACE

3/15/2015

21


Ex:

Sol:

Remark: The cross product is defined only for vectors in V3.

There is no corresponding operation for vectors in V2.


Sharjah University


OF SPACE


Theorem: For any vector , we have

1-

2-

Ex: Find and , if and

Exercises:- Compute for


Sharjah University

3a V

0a a

0 0a

u v v u 2u i j k 4 3v i j k


OF SPACE

1. 7,8, 7 , 3,4,2

2. 3 2 , 2 7

3. 2 4 , 4 2 3

a b

a i j b i k

a i j k b i j k

a b

3/15/2015

22


Theorem: For any vectors and in , and any scalars d

in the following holds:

Theorem: For nonzero vectors and in , if is the angle between and (0 ), then


Sharjah University

,a b c

1.

2.

3.

4. , ,

a b b a

d a b d a b a db

a b c a b a c

i j k j k i k i j

3V

3Va b

a b

sina b a b


OF SPACE


Corollary: Two nonzero vectors are parallel if and

only if

The Area of a Parallelogram: Consider the parallelogram

formed by the vectors and with an angle of between them. Then

Area of parallelogram =


Sharjah University

3,a b V

b

0a b

a

sina b a b

a

b


OF SPACE

3/15/2015

23


EX: Find the area of the parallelogram with two adjacent

sides formed by the vectors = 1, 2, 3 and = 4, 5, 6.

Sol:


Sharjah University

ba

Area of the parallelogram= a b

2 3 1 31 2 3

5 6 4 64 5 6

12 15 6 12 5 8 3 6 3 3,6, 3

i j k

a b i j k

i j k i j k

2 2 2

3,6, 3 3 6 3

9 36 9 54

a b


OF SPACE


EX: Find the area of the triangle determined by the three

points P(1,1,0), Q(0,-2,1) and R(1,-3,0).

Hints:

Finding the Volume of a Parallelepiped:-

The volume of parallelepiped determined by the vectors

and is the scalar triple product of the vectors and given by

where


Sharjah University

c

,a b

1Area of the triangle=

2a b

1 2 3

1 2 3

1 2 3

c

c c c

a b a a a

b b b

Volume of parallelepiped c a b


OF SPACE

3/15/2015

24


Ex: Find the volume of parallelepiped with three adjacent

edges formed by the vectors and

Sol:


Sharjah University

1,2,3 , 4,5,6a b

7,8,0 .c

Volume of parallelepiped c a b

1 2 3

1 2 3

1 2 3

7 8 0

c 1 2 3

4 5 6

2 3 1 3 1 27 8 0 7 12 15 8 6 12 0

5 6 4 6 4 5

7 3 8 6 21 48 27

The volume of parallelepiped c 27

c c c

a b a a a

b b b

V a b


OF SPACE


Exercises:-

1- Find the area of the triangle with vertices (0,0,0), (-4,-7,2),

and (-1, 5, -1).

2- Find the volume of the parallelpiped with three adjacent

edges formed by

3- Find the area of the parallelogram with two adjacent sides

formed by


Sharjah University

1, 6,0 , 1, 3,8 2, 7, 6u v and w


OF SPACE

6,5,0 5, 3,0u and v

3/15/2015

25

10.5: LINES AND PLANES IN SPACE

Let L be the line that passes through

the point P1(x1, y1, z1) and that is

parallel to the position vector

For any other point P(x, y, z) on the

line L, observe that the vector

will be parallel to , so that

for some scalar t.


Sharjah University

1 2 3, , .a a a a

1PPa

1PP ta


OF SPACE


Since

we have that

We obtain the parametric equations for a line

Provided none of or are zero. Prepared by Dr. F.G.A

Sharjah University

1 1 1 1, ,PP x x y y z z

1 1 1 1 1 2 3 1 2 3, , , , , , .PP x x y y z z ta t a a a ta ta ta

1 2,a a 3a


OF SPACE

3/15/2015

26


We can solve for the parameter in each of the three equations,

to obtain

We refer to the above equation as symmetric equations of

the line.

EX: Find equations for the line through the point (1, 5, 2) and

parallel to the vector 4, 3, 7. Also, determine where the line

intersects the yz-plane.

Sol:

Parametric Equations:


Sharjah University

1 1 1 4x x ta x t


OF SPACE


Symmetric Equations:

The line intersects the yz-plane where x = 0. Setting x = 0


Sharjah University

1 2

1 3

5 3

2 7

y y ta y t

z z ta z t

0 1 5 2

4 3 7

1 5 3 3 17For 5 5

4 3 4 4 4

1 2 7 7 1For 2 2

4 7 4 4 4

y z

yy y y

zz z z


OF SPACE

3/15/2015

27


So, the line intersects the yz-plane at the point

EX: Find equations for the line passing through the points

P(1, 2,1) and Q(5,3, 4).

Exercises:

1. Find parametric equations of the line through (-2,5,-6)

parallel to

2. Find symmetric equations of the line through (7,-3,2) and

parallel to

3. Find parametric equations of the line through (7,-5,7)

and (4,5,0). Prepared by Dr. F.G.A

Sharjah University


OF SPACE

2,4, 1 .

6, 9, 8 .


Definition: Let L1 and L2 be two lines in , with parallel

vectors and , respectively, and let be the angle between and .

1- The lines L1 and L2 are parallel whenever and are

parallel.

2- If L1 and L2 intersect, then

(a) the angle between L1 and L2 is (b) the lines L1 and L2 are orthogonal whenever and

are orthogonal.


Sharjah University

ba

ba

ba

ba


OF SPACE

3/15/2015

28


Definition: Nonparallel, Non intersection lines are called

Skew lines.


Sharjah University


OF SPACE


Ex: Show that the lines

are not parallel, yet do not intersect. (skew lines) SOL:

Parallel:- You can read from the parametric equations that

the vector parallel to L1 is , while

the vector parallel to L2 is .

Since is not a scalar multiple of , the vectors are not

parallel, and neither are the lines L1 and L2. Prepared by Dr. F.G.A

Sharjah University

1,2,2a

1

2

: 2 , 1 2 and 5 2

: 1 , 2 and 1 3

L x t y t z t

L x s y s z s

1, 1,3b 1 2 3, ,b b b b1 2 3, ,a a a a

ba


OF SPACE

3/15/2015

29


Intersection: If the line L1 intersect with the line L2, there is

a point (x,y,z) which would satisfy both lines equations. Setting the x-value (for both lines) equal, we get

Setting the y-value equal,

Solving this for t yields


Sharjah University

2 22 1 2 1 1

1 1

x t x tt s s t s t

x s x s

1 2 2 12 1 2 2 1

2 2

y t y tt s s t

y s y s

1 and 2 1 1 2 1 3 0 0

which implies that 1

s t s t t t t t

s


OF SPACE


Setting the z-components equal gives

but this is not satisfied when t = 0 and s = 1.

So, L1 and L2 are not parallel, yet do not intersect.

Ex: Let

Be two lines in . Determine whether L1 and L2 are parallel

or intersect or skew lines.


Sharjah University

1

2

: 1 4 , 5 4 and 1 4

: 2 8 , 4 3 and 5

L x t y t z t

L x s y s z s

3


OF SPACE

3/15/2015

30


The Equation of a Plane:

Let be a plane in and Is any point on . Let be any non zero

vector such that

To find an equation of the plane,

let P(x, y, z) represent any point in

the plane.

P and P1 are both points in the plane , so the vector lies in the plane and must be orthogonal to .


Sharjah University

3 1 1 1 1, ,P x y z

1 2 3, ,a a a a. ( orthogonalon )a a

a1 1 1 1, ,PP x x y y z z


OF SPACE


That is

Expanding the above expression, we get

We refer to this last equation as a linear equation in the

three variables x, y and z. This says that every linear

equation of the form

is the equation of a plane with normal

vector a, b, c. Prepared by Dr. F.G.A

Sharjah University

1 1 2 3 1 1 1

1 1 2 1 3 1

, , , , 0

0

PP a a a a x x y y z z

a x x a y y a z z


OF SPACE

3/15/2015

31


Ex: Find the plane containing the three points P(1, 3, 2),

Q(3,1, 6) and R(5, 2, 0). Sol:

Since both and lie in the plane, their cross product

is orthogonal to the plane and can be taken as

normal vector


Sharjah University

2 1 2 1 2 1

2 1 2 1 2 1

, , 3 1, 1 3,6 2 2, 4,4

, , 5 1,2 3,0 2 4, 1, 2

PQ x x y y z z

PR x x y y z z

PQ PRPQ PR

2 4 4 12 20 14

4 1 2

i j k

n PQ PR i j k


OF SPACE


With the point P(1, 3, 2) and the normal vector ,

an equation of the plane is

EX: Find the equation of the plane containing the point

(1,3,2) with normal vector

Sol:

EX: Find the equation of the plane containing the point

(-2,1,0) with normal vector Prepared by Dr. F.G.A

Sharjah University

n PQ PR

1 1 2 1 3 1 12 1 20 3 14 2 0

or

12 12 20 60 14 28 0 12 20 14 100

a x x a y y a z z x y z

x y z x y z

2 1 1 3 5 2 0.x y z

2, 1,5 .

3,0,2


OF SPACE

3/15/2015

32


Intersection of Planes:

We say that the two planes are parallel whenever their normal

vectors are parallel and the planes are orthogonal whenever

their normal vectors are orthogonal.

EX: Find an equation for the plane through the point (1, 4, 5) and parallel to the plane defined by 2x 5y + 7z = 12.

Sol:

A normal vector to the given plane is 2,5, 7.

Since the two planes are to be parallel, this vector is also

normal to the second plane. The equation of the plane is


Sharjah University


OF SPACE


EX: Find the line of intersection of the planes x + 2y + z = 3

and x 4y + 3z = 5. SOL:

------------------------

Substitute 3 in 1 we get

Taking y as the parameter (i.e., letting y = t), we obtain

parametric equations for the line of intersection:


Sharjah University

3 1 3z y

2 3 1x y z 4 3 5 2x y z

6 2 2 2 6 2 2y z z y

2 3 1 3 5 2 2 5 .x y y x y x y

3/15/2015

33

INTEGRATION TECHNIQUES


Ex:

1- Find the angle between the plane x+y+z=1 and x-2y+3z=1

2- Find symmetric equation for the line of intersection of these

two plane.

Remark: A line is parallel to a plane if and only if the direction

vector of the line is perpendicular to the normal vector of the

plane.

EX: Determine whether the line

and the plane are parallel.

Ex: find the equation of the plane that contains two lines


Sharjah University

: 2 10x z

: 5, 2 , 10 4L x y t z t

1

2

: 3 , 3 3 and 4

: 2 , 1 2 and 6 2

L x t y t z t

L x s y s z s


Distance From a Point to a Line:

The distance from a point to a line passes

through a point which is parallel to

a vector , is given by

Ex: Find the distance from the point Q(1, 2, 1) to the line

through the points P(2, 1,3) and R(2,1, 3). Sol

First, we have the position vectors


Sharjah University


OF SPACE

( , , )P x y z , ,P x y z

0P P ad

a

0 0 0 0, ,P x y z

a

0 0 0 0( , , )P x y z

a

3/15/2015

34


which gives us

Now, we have

Ex: Find the distance from the point P(1,1,5) to the line


Sharjah University


OF SPACE

: 1 , 3 and 2L x t y t z t


Distance From a Plane to a Point:

Let P be a point on a plane with

normal vector , then the distance

from any point S to the plane is

given by

EX: Find the distance from the point (1,3,0) and the plane

Sol:

First, find the point P in the plane which is easy to find from

the planes equation by taking the intercepts. Prepared by Dr. F.G.A

Sharjah University


OF SPACE

n

n

n PS nd PS

n n

3 5 2.x y z

3/15/2015

35


We take P to be y-intercept (x=0, z=0). We get


Sharjah University


OF SPACE

3 0 5 0 2. 2 0,2,0y y P

2 2 2

Fromplaneequation the normal vectors is 3,1, 5

3 1 5 9 1 25 35

1,3,0 , 0,2,0

1 0,3 2,0 0 1,1,0

1,1,0 3,1, 5 3 1 4

35 35 35

n

n

S P

PS

PS nd

n


Ex: Find the distance from the point (1,1,3) and the plane

EX: Find the distance from the point (2,0,1) and the plane

Ex: Find the distance between the two parallel planes:

Sol:

Note that the planes are parallel, since their normal vectors

2,3, 1 and 4,6, 2 are parallel, and the distance from the plane to every point in the plane must be the same.


Sharjah University


OF SPACE

3 2 6 6 0.x y z

2 2 4.x y z

1

2

: 2 3 6

: 4 6 2 8

x y z

x y z

1 2

3/15/2015

36


Accordingly, pick any point in , say P(0, 0, 6), and another

point in , say S(0,0,4), (we took the z-intercept for both

planes). We have


Sharjah University


OF SPACE

2 2 2

2

1

2 2 2

0 0,0 0,4 6 0,0, 2

2 3 1 4 9 1 14

The distance from the point (0, 0, 4) which is in

to the plane is then

0,0, 2 2, 3,1 0 0 2 2 2

14 14 142 3 1

which is the distance betwe

PS

n

PS nd

n

en the two planes.

1

2

Calculus 2 Chapter10 vectors in space

Documents

Transcript of Calculus 2 Chapter10 vectors in space