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Lesson 0３

Vector Equation

3A •  Vector Equation

Course III

2

Vector Equation of a Line

Vector equation of a line Line passing point A( ) parallel to the vector A

P

a!

d!

l

p!

a!

d!

dtap!!!

+=

O

dt!

t : parameter d!

: direction vector

Expression by components When , , and ),( 00 yxa =

!),( yxp =

! ),( yx ddd =!

),(),(),( 00 yx ddtyxyx +=

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Parametric Equation of a Line

A P

a!

d!

l

p!

O

dt!

Vector equation

),(),(),( 00 yx ddtyxyx +=

Parametric equation

⎭⎬⎫

+=

+=

y

x

dtyytdxx

0

0

x

y

O

l

A P

),( 00 yx),( yx

One more time ?

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Equation of a Line Passing A and B

)(a!

)(b!

Line passing two points A and B

.

A P

a!

d!

lp!

O

B

b!

abd!!!

−=

)( abtap!!!!

−+=Or

btatp!!!

+−= )1(

[Examples 3-1] Find the vector equation which passes through the points A=(1, 2) and B=(-2, 5)

Ans. )3,3()25,12(BA −=−−−=

!Since we have

)3,3()2,1(),( −+= tyx

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Straight Line and Normal Vector

Line Passing through Point A( ) and normal to the vector

a!

n!

A P

a!

lp!

O

n!

Since , we have n!!

⊥PA

0)( =−⋅ apn!!"

Expression by components

When , , and ),( 00 yxa =!

),( yxp =! ),( yx nnn =

!

0)()( 00 =−+− yynxxn yx

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Exercise

[Ex.3-1] Answer the following questions about the line which passes point A(3, 5) and is parallel to the vector

(1)  Find the vector equation. (2)  Find the parametric equation (3)  Derive the expression of the line by eliminating the parameter . Ans.

Pause the video and solve the problem by yourself.

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Answer to the Exercise

[Ex.3-1] Answer the following questions about the line which passes point A(3, 5) and is parallel to the vector . (1) Find the vector equation. (2) Find the parametric equation. (3) Derive the expression of the line by eliminating the parameter .

Ans.

Let the coordinates of the moving point on the line be (x, y). (1) (2) (3)

)4,2(−=d!

)4,2()5,3(),( −+= tyx

tytx 45,23 +=−=

5)3(2 −=− yx 112 +−=∴ xy

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Lesson 0３

Vector Equation

3B •  Vector Equation of a Circle

Course III

9

Vector Equation of a Circle　（１）

Vector equation Definition of a Circle P

r

p!

c!

x

C

O

yr=CP

Therefore rcp =−

!!

Making the square 22 rcp =−

!!

Therefore, from the definition of the scalar product

2)()( rcpcp =−⋅−!!!!

Case : Radius and center r c!

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Vector Equation of a Circle　（2）

Vector equation Circumference angle for a diameter is 90°

P

A

Case : Diameter AB, where A( ) and B( ) a!

b!

B 0)()( =−⋅− bpap!!!!

Next, we show the radius explicitly 22)()( rcacaca =−=−⋅−

!!!!!!Since

{ }0))(())((

)()()()(22=−−−=−−−−=

−⋅−−−=−⋅−

rcacpcacacp

cacacpcaap!!!!!!!!!!

!!!!!!!!!!

2))(( rcacp =−−!!!!

Therefore

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Example [Examples 3-2] Find the vector equation of the tangent to the circle at A( ). The radius of this circle is and the center is located at C( ).

Ans.

a!

r c!

Since the circumference angle for the diameter is 90°

0)()( =−⋅− caap!!!!

Since the radius is r22)()( rcacaca =−=−⋅−

!!!!!!

Substituting this, we have

{ }0))(())((

)()()()(22=−−−=−−−−=

−⋅−−−=−⋅−

rcacpcacacp

cacacpcaap!!!!!!!!!!

!!!!!!!!!!

Therefore 2))(( rcacp =−−

!!!!

A

p!

c!

x

C

O

P

a!

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Exercise [Ex3-2] What kind of figure does the following vector equation represent ?

Ans.

Pause the video and solve the problem by yourself.

323 =− ap!!

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Answer to the Exercise [Ex3-2] What kind of figure does the following vector equation represent ? Ans.

323 =− ap!!

This equation is rearranged as follows. 323 =− ap

!!

3323 =⎟

⎠

⎞⎜⎝

⎛−∴ ap!!

A circle with radius 1 and the center at the position

a!

32

A

p!

P

a!

32

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