13 Cartesian Space and Introduction to Vectors - Handout.pdf

The Cartesian Coordinate Space

and

Vectors in

R

Mathematics 54 - Elementary Analysis 2

The Three Dimensional Coordinate System

Recall:

Cartesian Coordinate System

To locate a point in space, three numbers are required. We represent any point in space by (a,b,c), an ordered triple of real numbers.

Definition

Coordinate Axes

To represent points in space:

These three coordinate axes determine thethree coordinate planes.

These three coordinate planes divide space into eight parts, calledoctants.

Example

Locate the following points in space.

1 A: (₋4, 3,₋5)

Distance Formula

Recall that inR2, the distance between two pointsP1=(x1,y1) and

P2=(x2,y2) is given by

d(P1,P2)= q

(x1−x2)2+(y1−y2)2

while inR3, we have

Theorem (Distance Formula)

The distance of between the points P1=(x1,y1,z1)and P2=(x2,y2,z2)is

d(P1,P2)= q

Example

1 The distance betweenA(1, 0,₋2) andB(9, 1, 2) is

d(A,B) ₌ p(1₋9)2₊₍₀₋₁₎2_+(−2−2)2

= p64+1+16=p81=9

2 _{The distance betweenQ(3,−5, 1) andR(2,−7, 3) is}

d(Q,R) = p(3−2)2_{+(−5−(−7))}2₊₍₁₋₃₎2

Midpoint Formula

InR2, the midpoint of the line segment connectingP1=(x1,y1) with

P2=(x2,y2) is given by

M₌³x1+x2

2 ,

y1+y2

2

´

InR3, we have

Theorem (Midpoint Formula)

The midpoint M of the line segment connecting the points P1=(x1,y1,z1)

and P2=(x2,y2,z2)is

M₌³x1+x2

2 ,

y1+y2

2 , z1+z2

2

Example

1 _{The midpoint of the line segment connecting pointsA(4,−5, 1) and}

R(2,−7,−3) is

M₌

µ₄

+2 2 ,

−5₋7 2 ,

1₋3 2

¶

=(3,−6,−1)

2 _{The midpoint of the line segment connecting the pointsP(3,−2, 9)}

andQ(−5, 2,−7) is

M₌

µ₃

−5 2 ,

−2₊2 2 ,

9₋7 2

¶

Spheres

Definition

Asphereis the set of all pointsP(x,y,z) whose distance from the pointCis r_>0. The pointCis called thecenterof the sphere, andritsradius.

Consider a sphere centered atC(h,k,l) with radiusr.

Thus,

d(P,C)=r =

q

(x−h)2_+(y−k)2_+(z−l)2

Spheres

Equation of a Sphere

The equation of the sphere centered atC(h,k,l) with radiusris given by

(x₋h)2₊(y₋k)2₊(z₋l)2₌r2

Specifically, the sphere centered at the origin with radiusrhas equation

Examples

1 Find the equation of the sphere centered at (3, 1,₋5) with diameter of

8 units.

Solution:

The radius of the sphere is 4, which is half of the diameter. Hence it equation is

Examples

2 _{Show that}x2₊y2₊z2₋4x₊2y₊6z_{−11=0 is a sphere, and find its} center and radius.

Solution:

By completing the squares, we have

(x2−4x+4)+(y2+2y+1)+(z2+6z+9) = 11+4+1+9

(x₋2)2₊(y₊1)2₊(z₊3)2 ₌ 25

Examples

3 _{Find the equation of the sphere centered atA(3,−5, 1) and contains}

the pointB(1, 0,−1).

Solution:

The radiusrof this sphere is given by the distance between the pointsA andB, i.e.,

r = d(A,B)=p(3−1)2_+(−5−0)2_{+(1−(−1))}2

= p4₊25₊4₌p33

Hence, this sphere will have an equation

Vectors

A vector is a quantity that has both magnitude and direction. Examples are displacement, velocity, force, etc.

Consider the directed line segment from the origin going to the point (a,b,c).

This is position representation of the vector_〈a,b,c_〉.

z

y

x a

b c

(a,b,c)

~v

Remark

1 _{If the position representation of}~_{vhas terminal point (a,b,c) thena,b} andcare called thecomponentsof~v.

2 A vector~vcan be represented by any directed line segment that has the same magnitude and direction as of~v.

z

y

x

~v

~u

Here we say that~vand~uareequal(orequivalent) and write~v₌~u.

On that note, we can define a vector from an initial point to another point.

Remark

4 _Let~_{vbe the vector from the pointP(a,b,c) going to the pointQ(d,e,f),}

then~v_{= 〈}d₋a,e₋b,f₋c_〉. Often times, we write this vector as−→PQ.

z

y

x

P(a,b,c)

Q(d,e,f)

−→

Example

1 _{The vector from the point (4, 1,−3) to the origin is given by}

~v _{= 〈}0−4, 0−1, 0−(−3)〉 = 〈−4,−1, 3〉

2 The vector from the point (2,₋5, 1) to the point (0, 2,₋4) is given by

Norm of a Vector

Definition

The magnitude (or length) of the vector~v_{= 〈}a,b,c_〉, called itsnorm, is given by

k~vk =pa2_+b2_+c2

Note

Example

Find the magnitude of the following vectors.

1 _〈3,₋5,₋8_〉

2

¿₁

2, 3 2,−

5 2

À

3

¿

−p1 10, 2 p 10, 1 p 2 À Solutions:

1 _{k〈3,−5,−8〉k =}p_9+25+64=p₉₈

2 ° ° ° ° ¿₁ 2, 3 2,−

5 2 À° ° ° °= r 1 4+ 9 4+ 25 4 = p 35 2 3 ° ° ° ° ¿

Remarks

1 _{A vector of length or magnitude 1 is called aunit vector.}

2 If_k~v_{k =}0 then~v₌~0_{= 〈}0, 0, 0_〉, and is called thezero vector.

Note

Directional Angles

Thedirection anglesof a nonzero vector~vare the anglesα,β, andγ(in the interval [0,π]) that it makes with the positivex,y, andz-axes.

The cosines of these direction angles,α,β, andγare called thedirectional cosinesof the vector~v.

So for~v_{= 〈}a,b,c_〉, we have

Directional Angles

Example

Find the directional angles of the vector~v_{= 〈}2,₋1, 3_〉.

Solution:

Since_k~v_{k =}p4₊1₊9₌p14 , then

cosα=p2

14, cosβ= −1 p

14 and cosγ= 3 p

14

and so

α=cos−1

µ ₂

p

¶

, β₌cos−1

µ

−1 p

¶

and γ₌cos−1

µ ₃

p

Vector Operations

When adding two vectors

z

y

x

~v ~u

~u

−−−→

Vector Operations

And when subtracting

z

y

~v ~u

~u

−−−→

Vector Operations

So if~A_{= 〈}a1,a2,a3〉and~B= 〈b1,b2,b3〉

Definition (Vector Addition / Subtraction)

~A_±~B_{= 〈}a1±b1,a2±b2,a3±b3〉

~A

~B

~C

~D

~A₊~B

~

C₋~D

Vector Operations

Let~A_{= 〈}a1,a2,a3〉andc∈R

Definition (Scalar Multiplication)

c~A_{= 〈}ca1,ca2,ca3〉

z

y

~v −→

2v

Remark

1 The length ofc~A_{= |}c_{| · k}Ak

kc~A_{k =}

q

(ca1)2+(ca2)2+(ca3)2

= |c_|

q

a2

1+a22+a23

kc~A_{k = |}c_{| · k}A_k

2 _{Ifc>0, thenc}~_{Ahas the same direction as}~_A.

3 And ifc_<0, thenc~Ahas the opposite direction as~A.

Example

Given~A_{= 〈}1, 2, 3_〉and~B_{= 〈−}4, 5,₋6_〉

1 ~_A+~_{B= 〈1, 2, 3〉 + 〈−4, 5,−6〉 = 〈−3,7,−3〉}

2 ₂~_A−~_{B= 〈2, 4, 6〉 − 〈−4, 5,−6〉 = 〈6,−1,12〉}

3 1

2~B+ 3 2~A=

¿

−2,5 2,−3

À

+

¿₃

Now let us consider the following unit vectors:

ˆ

ı:= 〈1, 0, 0〉

ˆ

:= 〈0, 1, 0〉 ˆ

k:= 〈0, 0, 1〉

z

y

x

Note that any vector~v= 〈a,b,c〉can be written as

~v = a〈1, 0, 0〉 +b〈0, 1, 0〉 +c〈0, 0, 1〉 = aˆı₊bˆ₊ckˆ

For this reason, ˆı, ˆand ˆkare called thestandard basis vectorsforR3.

Example

Given a vector~v6=~0, we cannormalize~v, i.e., transform it into a unit vector having the same direction as~v, by:

~v˜= 1 k~v_k~v

Note

1 °°~v˜ ° °= ° ° ° ° 1 k~v_k~v

° ° ° °= ¯ ¯ ¯ ¯ 1 k~v_k

¯ ¯ ¯ ¯k~

v_{k =}1 (since_k~v_{k >}0)

2 k~vk~v˜₌~vand sincek~vk >0, then~v˜and~vhave the same direction.

3 _{It is easy to see that the components of}~v˜_{are the directional cosines} of~v, which implies that

Example

Find a unit vector~vthat has the same direction as the vector−→PQ, where P(3, 2, 1) andQ(0, 5,−1).

Solution: −→

PQ_{= 〈−}3, 3,₋2_〉

° ° ° −→ PQ ° ° °= p

9+9+4=p22

Hence,~v_=p1

22〈−3, 3,−2〉 =

Exercises

1 Locate the following points

1 (2, 4,₋1)

2 (−3, 1,−2)

3 (7,₋2,₋6)

4 (−2,−3,−4)

2 _{LetA(1,−5, 2),B(3, 2,−4) andC(−4, 1, 3). Find}

1 the midpoint ofDCwhereDis the midpoint ofAB.

2 a point in thez-axis that is equidistant to bothAandB.

3 the sphere centered atC, containingB.

3 _Define~v_{as the vector from (3, 1,−2) to (1, 5, 2). Findk}~v_{kand its} directional cosines.

4 If~u_{= 〈}2,₋3, 1_〉,~v_{= 〈}1, 0,₋1_〉and~w_{= 〈−}1, 3,₋2_〉, find: