Lecture 2. Vectors and GeometryUniversity of British Columbia, Vancouver

Yue-Xian Li

January 17, 2019

1

2.1 Scalars vs Vectors

Scalar: Any number in R is referred to as a scalar, where Ris the set of all real numbers. A scalar is supposed to have adefined magnitude.

Note that: the concept can be extended to refer to any num-ber in C as a complex scalar, where C is the set of all complexnumbers.

Vector: Any quantity determined by two or more scalars ar-ranged in predetermined order. A vector is supposed to haveboth a defined magnitude and a direction.

Eg 2.1.1: A point in R2 (2D space) is a 2D vector.

0 a

(a,b) b

y

x

(a,b,c)

b

x

y

z

0

a

c

Eg 2.1.2: A point in R3 (3D space) is a 3D vector.

2

Eg 2.1.3: Midterm performance of a student in Math 152can be defined as a 19D vector.

(hw1, · · · , hw11, lb1, · · · , lb6,md1,md2) ∈ R19

where hw1, · · · , hw11 are the 11 homework assignment marks,lb1, · · · , lb6 the 6 labwork marks, andmd1,md2 the 2 midtermexam marks.

Remarks:

(i) Each scalar in a vector is called a component (or entry),e.g. the scalar 2 in vector (3, 1, 2) is the 3rd component ofthe vector.

(ii) The number of components in a vector = dimension of thevector.

(iii) The order/position of each component in a vector mustbe consistent among all vectors, e.g. if (2,−1) is a vectordescribing a point in a 2D space, 2 (but not −1) must bethe x coordinate of the point.

3

In physics and geometry:

A vector is referred to as a quantity with both a magnitudeand a direction.

Remarks:

• There is one special exception, the vector ~0 = (0, 0) ∈ R2

is a vector with magnitude 0 and no direction.

Question: How to determine the magnitude and directionof a vector?

Answer: Given any vector ~x = (x, y) 6= (0, 0) ∈ R2, drawan arrowed line connecting (0, 0) to (x, y) (see figure below),length of the line represents the magnitude while the arrowdirection represents the direction of the vector.

y

x0

x =(x,y)

4

Eg 2.1.4: ~a = (1, 1) (or [1, 1]). Its direction is 45◦ (seefigure) and its magnitude, represented by double vertical bars

||~a|| =√

12 + 12 =√

2,

where ||~a|| defines the magnitude/length of the vector~a. (Notethat it differs from the notation |a| which denotes the magni-tude or absolute value of a scalar a.)

x

1a =(1,1)

2

a

1 3

y

0

a

• Usually, (1, 1) can be regarded as the location of the arrowhead relative to (0, 0) when the vector is drawn as an arrowconnecting (0, 0) and (1, 1).

• But all arrows with identical direction and magnitude rep-resent the same vector irrespective of the location of thearrow tail (see figure!)

• Question: what is the meaning of (1, 1) when the vectoris represented by the red arrow in the figure?Answer: it represents the location of the arrow headrelative to the location of the arrow tail!

5

Summary:

To graphically determine the direction and magnitude of anyvector

~a = (a1, a2) ∈ R2,

one picks any starting point T as the “tail” location, move a1units horizontally and a2 units vertically to locate the endingpoint H as the “head”, draw an arrowed line from T to H .The arrow would point to the direction of ~a and the length ofthe line would represent its magnitude.

0

H

T1

xa

a2

a1 a2a=( , )

y

6

2.2 Vector addition and scalar multiplication

Let ~a = (a1, a2), ~b = (b1, b2) be vectors in R2,and c, d be scalars. Then,

(i) ~a±~b = (a1 ± b1, a2 ± b2);i.e. adding/subtracting the corresponding components;

(ii) c~a = (ca1, ca2);i.e. multiplying each component by the scalar;

(iii) −~a = (−a1,−a2);(iv) ~a− ~a = (0, 0) = ~0,

where ~0 is a vector with zero-length and no direction!

7

Properties of addition/scalar multiplication

Let ~a, ~b, ~c, ~0 be vectors in Rn (n ≥ 2),and d, e be scalars. Then,

1. ~a +~b = ~b + ~a (commutative);

2. ~a + (~b + ~c) = (~a +~b) + ~c (associative);

3. ~a +~0 = ~a (existence of zero);

4. ~a + (−~a) = ~a− ~a = ~0 (existence of negative);

5. d(~a +~b) = d~a + d~b (distributive with scalar product);

6. (d + e)~a = d~a + e~a (distributive with scalar product);

7. 1~a = ~a.

8

Geometric meaning of vector addition/subtraction

Eg 2.2.1: Given that ~a = (2, 1), ~b = (1, 2). Demonstrate

the geometric meaning of: (1) ~c = ~a +~b; (2) ~c = ~a −~b; (3)~c = 2~a, 1

2~a; (4) ~c = −~a.

Answer: (1) ~c = ~a +~b = (2, 1) + (1, 2) = (3, 3).

a0

a

x

b

2

b

1

c

3

a2 b

=+

3

1

y

• Plot ~a and ~b at identical starting point, they form twosides of a parallelogram (pgram). ~c = ~a+~b is the diagonalvector of the pgram starting from the same point.

• Connecting ~a and~b one-by-one, tail-to-head, irrespectivelyof order, the vector that connects the tail of the 1st andthe head of the last represents the sum.

•When adding more than two vectors, one needs to connectthem one-by-one, tail-to-head, irrespectively of order, thevector that connects the tail of the 1st and the head of thelast vector represents the sum of all.

9

(2) ~c = ~a−~b = (2, 1)− (1, 2) = (1,−1).

−−1

a

1b

0

b1

c2

a

y

b= −

c

x

a b

2

=

• Plot ~a and ~b at identical starting point (i.e. join the tails

of the two), they form two sides of a triangle. ~c = ~a −~bis the 3rd side of the triangle connecting the heads of thetwo, starting from the head of ~b to that of ~a.

• Plot ~a and ~b at identical ending point (i.e. join the heads

of the two), they form two sides of a triangle. ~c = ~a−~b isthe 3rd side of the triangle connecting the tails of the two,starting from the tail of ~a to that of ~b.

10

(c) ~c = 2~a = (4, 2), ~c = 12~a = (1, 12).

2 3

a

4

a

5

c =2

a

y

ac =1/2

0

1x

2

1

• ~c = 2~a is a vector in the same direction as ~a but withdoubled length.

• ~c = 12~a is a vector in the same direction as ~a but with half

length.

• ~c = γ~a (γ 6= 0 is scalar), a vector in the same direction as~a but with a length scaled by a factor γ.

• Scalar multiplication of a vector does not change its direc-tion but only its length.

11

(d) ~c = −~a = (−2,−1).

1 a

−1

c

−2

a = −

y

x0 1 2

−1

• ~c = −~a has the same length as ~a but with direction re-versed.

• The negative of a vector is a vector of identical length butopposite in direction.

12

Eg 2.2.2: Equation of a circle of radius 1 centred at ~c =(1, 2).

2

c

y

x =(x,y)

0

1

− cx2

x1

Ans: Pick any point on the circle ~x = (x, y), ~x − ~c definesthe radial vector (red) starting from the centre of the circlepointing to that point. Thus, its length must be equal to theradius, i.e. 1. Therefore,

||~x− ~c|| = 1

describes the set of all points ~x whose distance from the centreis equal to 1. This is exactly the circle itself.

• Note that ||~x− ~c|| =√

(x− 1)2 + (y − 2)2 = 1 becomesthe equation of circle (x − 1)2 + (y − 2)2 = 12 that weknow previously.

• However, ||~x − ~c|| = 1 can also describe geometricallysimilar objects in higher dimensions. If ~x,~c ∈ R3, thenit describes a sphere of radius 1 centred at ~c. If ~x,~c ∈Rn (n > 3), then it describes a hypersphere of radius 1centred at ~c in n-dimensional space.

13

2.3 Vectors in orthogonal coordinates

In orthogonal coordinate systems, each vector can be ex-pressed as a linear combination of unit vectors representingthe directions of the orthogonal axes, also referred to as thebasis vectors.

In R2, the basis vectors are

i = ~e1 = (1, 0), j = ~e2 = (0, 1).

Thus, any vector in R2 can be expressed as

~a = (a1, a2) = a1(1, 0) + a2(0, 1) = a1i + a2j,

where a1, a2 are the coordinates of ~a in orthogonal basis i, j.

x

i

j

a

a1

a2

1

y

10

14

In R3, the basis vectors are

i = ~e1 = (1, 0, 0), j = ~e2 = (0, 1, 0), k = ~e3 = (0, 0, 1).

Thus, any vector in R3 can be expressed as

~b = (b1, b2, b3) = b1(1, 0, 0) + b2(0, 1, 0) + b3(0, 0, 1)

= b1i + b2j + b3k.

where b1, b2, b3 are the coordinates of ~b in orthogonal basisi, j, k.

z

i j

k

b

b1

2b

b3

1

x

y1

1

15

2.4 Inner/dot product

Def: Let ~a = (a1, a2, · · · , an), ~b = (b1, b2, · · · , bn), ~c bevectors in Rn and that d be a scalar. The inner/dot productis defined as

~a ·~b ≡ a1b1 + a2b2 + · · · + anbn,

which is a scalar.

Important properties:

(i) ~a ·~b = ~b · ~a (commutative);

(ii) ~a · ~a = ||~a||2 = a21 + a22 + · · · + a2n;

(iii) ~a · (~b + ~c) = ~a ·~b + ~a · ~c (distributive);

(iv) (d~a) ·~b = ~a · (d~b) = d(~a ·~b) (commutative with scalar);

(v) ~0 · ~a = ~a ·~0 = ~0 (existence of zero);

∗(vi) ~a ·~b = ||~a||||~b|| cos θ, θ=smallest angle between ~a, ~b;

(vii) ~a ·~b = 0 iff ~a = ~0 or ~b = ~0 or ~a ⊥ ~b.

All results above can be proved using the definition. (vi) isequivalent to the Law of Cosines. Let’s prove this result whichcan also be regarded as a proof of the Law of Cosines.

16

Proof: ~a ·~b = ||~a||||~b|| cos θ.

Law of Cosines:

c2 = a2 + b2 − 2ab cos θ.

a c

b

θ

Based on the definition and other properties of inner product

||~c||2 = ~c · ~c = (~a−~b) · (~a−~b) = ||~a||2 + ||~b||2 − 2~a ·~b.

Based on Pythagorean Theorem (see figure)

cb

b cosθ a b cosθ−

a

bsin

θ

b

a

a bc= −

θθ

||~c||2 =(||~b|| sin θ

)2+(||~a|| − ||~b|| cos θ

)2= ||~a||2 + ||~b||2 − 2||~a||||~b|| cos θ.

17

The last expression above is actually the Law of Cosines. Bycomparing the two expressions for ||~c||2, we conclude

~a ·~b = ||~a||||~b|| cos θ.

Eg 2.4.1: Given ~a = (2, 3), ~b = (1,−3) ∈ R2. Calculate

(a) 2~a + 4~b;

(b) ~a ·~b;

(c) ||~a||, ||~b||;(d) cos θ, θ.

Ans:

(a) 2~a + 4~b = 2(2, 3) + 4(1,−3) = (8,−6);

(b) ~a ·~b = (2, 3) · (1,−3) = 2− 9 = −7;

(c) ||~a|| =√

22 + 32 =√

13, ||~b|| =√

12 + (−3)2 =√

10;

(d) cos θ = ~a·~b||~a||||~b||

= −7√130, θ ≈ 127.875◦ ≈ 2.23184 rad.

18

Eg 2.4.2: Find the angle between vectors connecting thecentre of a cube to two neighbouring vertices.

Ans: Consider a cube of length 2 centred at ~0.

2

θ

b

a

x

=(1,−

1,1)

=

3

b

= 3

=(1,1

,1)

z

θ a

y

Pick two vertices located at:

~a = (1, 1, 1), ~b = (1,−1, 1).

Applying the inner product law,

cos θ =~a ·~b||~a||||~b||

=1

3⇒ θ = cos−1

1

3≈ 1.231 rad ≈ 70.5◦.

Using the Law of Cosines, we get the same result

22 = (√

3)2 + (√

3)2 − 2(√

3)2 cos θ ⇒ cos θ =2

6=

1

3.

19

2.5 Dot product and projection

Def: Projection of ~a on ~b, denoted by Proj~b~a, is defined as

a vector in direction of ~b with a length that is equal to the“shadow” of ~a on ~b (see figure).

baProj

b

a

θ

Thus,

Proj~b~a = (“shadow” length of ~a on ~b)(direction of ~b)

= (||~a|| cos θ)~b

||~b||= (~a ·~b)

~b

||~b||2.

This gives the universal projection formula,

Proj~b~a = (~a ·~b)~b

||~b||2.

20

If ~b = ~u is a unit vector (i.e. ||~u|| = 1), then

Proj~u~a = (~a · ~u)~u.

Eg 2.5.1: Given ~a = (2, 3), ~b = (1,−3) ∈ R2, find Proj~a~b.

Ans: Using the projection formula

Proj~a~b = (~a ·~b) ~a

||~a||2=−7

13~a =

(−14

13,−21

13

).

Eg 2.5.2: Force experienced by a pendulum parallel andperpendicular to the direction of motion.

Ans:

~F|| = Proj~u||~F = (~F · ~u||)~u||

= −mg sin θ~u||;

F

jF=−mg

m

uu

F

θ

θ

~F⊥ = Proj~u⊥~F = (~F · ~u⊥)~u⊥

= −mg cos θ~u⊥.

21

2.6 Dot product, area of parallelograms (pgrams),and matrix determinants

Let A=area of the pgram with sides ~a, ~b.

a

b

aba

Proj

2θ−

ππ

θ

A = (base)(height) = ||~a||||~b|| sin θ = ||~a||||~b|| cos(π

2− θ).

Let ~a⊥ be a vector perpendicular to ~a with identical length,i.e. ~a⊥ · ~a = 0 and ||~a⊥|| = ||~a||. Based on the figure,

A = ||~a||||~b|| cos(π

2− θ) = ~a⊥ ·~b,

note that π2 − θ is the angle between between ~a⊥ and ~b.

If ~a = (a1, a2), ~b = (b1, b2), then ~a⊥ = (−a2, a1). Thus,

A = ~a⊥ ·~b = a1b2 − a2b1.

22

Def: A matrix is an array of scalars typically enclosed bysquare or round brackets.

Eg 2.6.1:

2 1−3 01 4

or

2 1−3 01 4

is a 3 × 2 matrix

with 3 rows and 2 columns.

Def: A vector is a column or a row of scalars, or a specialmatrix with only one column or one row.

Def: A scalar is a 1×1 matrix with one row and one column.

Def: For a 2× 2 matrix, the determinant is defined as

.1b 2b

+−

a1 2b a2 1

a− b

2a1

1b 2bdet

a2a1

Therefore, the area of a pgram with sides ~a, ~b is

A = ~a⊥ ·~b =

∣∣∣∣det

[a1 a2b1 b2

]∣∣∣∣ = abs

(∣∣∣∣ a1 a2b1 b2

∣∣∣∣) .23

Eg: Calculate the area of the pgram sided with ~a = (−1, 2)

and ~b = (1, 3).

Ans:

A = abs

(∣∣∣∣ −1 21 3

∣∣∣∣) = abs(−3− 2) = 5.

Similarly, one can calculate the volume of a parallelepiped(ppiped) with edges ~a = (a1, a2, a3), ~b = (b1, b2, b3), and~c = (c1, c2, c3) in 3D space.

b

c

a

V =

∣∣∣∣∣∣det

a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ = abs

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ .

(Proof will come later!)

24

Def: For a 3× 3 matrix, the determinant is defined as:

(1) in row expansion formula

1c 2c 3c

a1 1b 2b b3

a1 a2 a3

1c 2c 3c

1b 2b b3

a1 a2 a3

1c 2c 3c

a2 1b 2b b3

a1 a2 a3

1c 2c 3c

a3− +

a2b3

3c

b

c

1

1

a12b b3

2c 3ca3

b1 b2

c2c1

− +

det1b 2b b3

a1 a2 a3

1 2 3c c c

1b 2b b3

a1 a2 a3

(2) in diagonal product formula

− − − +++ +−

3

b3

3c

a1 a2

1b 2b

1c 2c

a1 a2

1b 2b

1c 2c

1b 2b b3

a1 a2 a3

1c 2c 3c

det1b 2b b3

a1 a2 a3

1 2 3c c c

1b 2b b3

a1 a2 a3

1c 2c 3c

a

25

Eg 2.6.2: Calculate the volume of the ppiped with edges

(1) ~a = (3, 0, 0), ~b = (2, 2, 0), ~c = (1,−1,−1);

(2) ~a = (1, 2, 3), ~b = (2, 3, 4), ~c = (1, 1, 1);

(3) ~a = (1, 2, 3), ~b = (2, 1, 2), ~c = (1, 1, 1).

Ans:

(1)

V1 = abs

∣∣∣∣∣∣3 0 02 2 01 −1 −1

∣∣∣∣∣∣ = abs

(3

∣∣∣∣ 2 0−1 −1

∣∣∣∣) = |−6| = 6.

(2)

V2 = abs

∣∣∣∣∣∣1 2 32 3 41 1 1

∣∣∣∣∣∣ = |3 + 6 + 8− 9− 4− 4| = 0.

This is because ~c = ~b−~a which means that the 3 vectorsare in the same plane. Note that the determinant is zeroif two rows or columns are identical or one is the linearcombination of the other two.

(3)

V3 = abs

∣∣∣∣∣∣1 2 32 1 21 1 1

∣∣∣∣∣∣ = |1 + 6 + 4− 3− 2− 4| = 2.

26

2.7 Cross product (between 3D vectors)

Def: Let ~a = (a1, a2, a3), ~b = (b1, b2, b3) be vectors in R3.The cross product

~a×~b =

∣∣∣∣∣∣i j ka1 a2 a3b1 b2 b3

∣∣∣∣∣∣ = i

∣∣∣∣ a2 a3b2 b3

∣∣∣∣− j ∣∣∣∣ a1 a3b1 b3

∣∣∣∣ + k

∣∣∣∣ a1 a2b1 b2

∣∣∣∣= i(a2b3 − a3b2)− j(a1b3 − a3b1) + k(a1b2 − a2b1)

is a vector perpendicular to the plane span by ~a and ~b with adirection determined by the right-hand-rule (RHR).

b

a

a bx

θ

27

Important properties: Let ~a, ~b, ~c be vectors in R3, d be a scalar.

Although we do not present proof of each of these properties, we can regard

them as results rigorously proven and can be used in the proof of other

theorems. The ones marked by ∗ will be proved in the following text.

(1) ~a, ~b, ~a×~b are all vectors in R3.

∗(2) ~a×~b is ⊥ to ~a, ~b following the RHR,

i.e. ~a · (~a×~b) = 0 = ~b · (~a×~b).∗(3) ||~a×~b|| = ||~a||||~b|| sin θ = the area of pgram with sides ~a, ~b;

θ=the smallest angle between ~a, ~b.

(4) In a RH coordinate system:

i× j = k, j × k = i, k × i = j.

(5) ~a×~b =(||~a||||~b|| sin θ

)~u⊥, where

||~u⊥|| = 1, ~u⊥ ⊥ ~a, ~b following RHR.

(6) ~a×~b = ~0 iff ~a = ~0 or ~b = ~0 or ~a ‖ ~b.

(7) ~a×~b = −~b× ~a.

(8) (d~a)×~b = ~a× (d~b) = d(~a×~b).

(9) ~a× (~b + ~c) = ~a×~b + ~a× ~c.

∗(10) ~a · (~b× ~c) = (~a×~b) · ~c = ~b · (~c× ~a) =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣.

28

Proof of properties (10) and (2):

For (10):

~a · (~b× ~c) = ~a ·

∣∣∣∣∣∣i j k

b1 b2 b3c1 c2 c3

∣∣∣∣∣∣= (a1i + a2j + a3k) ·

[i

∣∣∣∣ b2 b3c2 c3

∣∣∣∣− j ∣∣∣∣ b1 b3c1 c3

∣∣∣∣ + k

∣∣∣∣ b1 b2c1 c2

∣∣∣∣]

= a1

∣∣∣∣ b2 b3c2 c3

∣∣∣∣− a2 ∣∣∣∣ b1 b3c1 c3

∣∣∣∣ + a3

∣∣∣∣ b1 b2c1 c2

∣∣∣∣ =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ .Similarly, one can show that

(~a×~b) · ~c = ~b · (~c× ~a) =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ .For (2),

~a · (~a×~b) =

∣∣∣∣∣∣a1 a2 a3a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ = 0 (two identical rows!).

Similarly,

~b · (~a×~b) =

∣∣∣∣∣∣b1 b2 b3a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ = 0 (two identical rows!).

Thus, ~a and ~b are both ⊥ to ~a×~b!

29

∗Proof (3): ||~a×~b|| = ||~a||||~b|| sin θ. We use Lagrange’s Identity:

||~a×~b||2 = ||~a||2||~b||2 − (~a ·~b)2,

which is proved as follows.

LHS =||~a×~b||2 = (a2b3 − a3b2)2 + (a1b3 − a3b1)2 + (a1b2 − a2b1)2

=a22b23 + a23b

22 + a21b

23 + a23b

21 + a21b

22 + a22b

21

− 2 (a2a3b2b3 + a1a3b1b3 + a1a2b1b2)

RHS =||~a||2||~b||2 − (~a ·~b)2

=(a21 + a22 + a23)(b21 + b22 + b23)− (a1b1 + a2b2 + a3b3)

2

=a22b23 + a23b

22 + a21b

23 + a23b

21 + a21b

22 + a22b

21 + a21b

21 + a22b

22 + a23b

23

−[a21b

21 + a22b

22 + a23b

23 + 2 (a1a2b1b2 + a1a3b1b3 + a2a3b2b3)

]=a22b

23 + a23b

22 + a21b

23 + a23b

21 + a21b

22 + a22b

21

− 2 (a1a2b1b2 + a1a3b1b3 + a2a3b2b3)

Calculations above clearly shows that LHS = RHS, i.e. for ~a, ~b ∈ R3

||~a×~b||2 = ||~a||2||~b||2 − (~a ·~b)2.

Now, substitute ~a ·~b = ||~a||||~b|| cos θ into Lagrange’s Identity above,

||~a×~b||2 = ||~a||2||~b||2 − ||~a||2||~b||2 cos2 θ

= ||~a||2||~b||2(1− cos2 θ)

= ||~a||2||~b||2 sin2 θ ⇒

||~a×~b|| = ||~a||||~b|| sin θ.

30

Theorem: The volume of a ppiped with edges ~a, ~b, ~c is

V =∣∣∣~a · (~b× ~c)∣∣∣ =

∣∣∣∣∣∣det

a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ = abs

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ .

Proof:

θ

θ

c

b cx

b cx

cos

a

b

a

V = (base area)(height) = ||~b× ~c||||~a|| cos θ = ~a · (~b× ~c).

31

Eg 2.7.1: Let ~a = (1, 3,−2), ~b = (−1, 2, 3), ~c = (1, 1, 1). Find

(1) area of the pgram with sides ~a and ~b;

(2) angle between ~a and ~b;

(3) volume of the ppiped with edges ~a, ~b, ~c.

Ans:

(1) ~a×~b =

∣∣∣∣∣∣i j k

a1 a2 a3b1 b2 b3

∣∣∣∣∣∣ =

∣∣∣∣∣∣i j k

1 3 −2

−1 2 3

∣∣∣∣∣∣= 13i− (1)j + 5k = (13,−1, 5).

A = ||~a×~b|| =√

132 + (−1)2 + 52 =√

195 ≈ 13.96.

(2) cos θ =~a ·~b||~a||||~b||

=−1 + 6− 6√

14√

14=−1

14.

θ = cos−1(−1

14

)≈ 1.64 rad ≈ 94.1◦.

Question: Why not use the formula ||~a×~b|| = ||~a||||~b|| sin θ here?

Answer: Because sin(θ) = sin(π − θ), thus sin−1(0.5) = π6 and

π − π6 = 5π

6 . We would not know which one to choose. E.g.

θ = sin−1

(||~a×~b||||~a||||~b||

)= sin−1

√195

14≈ 85.9◦

This is not the right answer. The right answer should have been

θ = 180◦ − 85.9◦ = 94.1◦!

(3) V =∣∣∣~a · (~b× ~c)∣∣∣ = abs

∣∣∣∣∣∣1 3 −2

−1 2 3

1 1 1

∣∣∣∣∣∣ = 2+9+2+4−3+3 = 17.

32

2.8 Lines, curves, planes expressed in vector forms

2.8.1 Lines in 2D space

(1) Point-direction formula: Find the equation of line L ifone point on it and its direction are given. Let ~p = (p1, p2)

be the point and ~l = (l1, l2) be the direction.

p

y

x

t>0

t<0t=0

0L

− px

x

l

Based on the graph, any point ~x = (x, y) on L should satisfy

~x− ~p = t~l, (parametric form, t ∈ R is a parameter.)

To put it in words: the difference between any point ~xon L and the point ~p lies in the same direction as ~l.

Thus,

~x = t~l + ~p.

33

Remarks:

•When t = 0, ~x = ~p.

•When t > 0 and as t increases, ~x moves away from ~pfollowing the ~l direction.

•When t < 0 and as t decreases, ~x moves away from ~pfollowing the −~l direction.

• If ~p = ~0, then line L passes through the origin.

34

(2) Point-normal formula: Find the equation of line L if onepoint on it and its normal direction are given. Let ~p = (p1, p2)be the point and ~n = (n1, n2) be the normal direction.

p

x

L

x p−

n

l

0

y

x

Based on the graph, any point ~x = (x, y) on L should satisfy

~n · (~x− ~p) = 0.

To put it in words: the difference between any point ~xon L and the point ~p must be perpendicular to ~n.

Remarks:

• The point-normal formula does not involve a parameter.

• Given~l = (l1, l2), one knows that ~n = (l2,−l1) or (−l2, l1).

• Given ~n = (n1, n2), one knows that ~l = (n2,−n1) or(−n2, n1).

35

Eg 2.8.1: Express the equation of line L defined by y =3x + 7 in two vector forms.

Ans: One needs a point on L: pick ~p = (0, 7) for simplicity.

One needs the direction of L and its normal, given the slopem = 3,

~l = (1, 3), ⇒ ~n = (3,−1).

Thus, in point-direction formula

~x = t~l + ~p = t(1, 3) + (0, 7), t ∈ R.

Or {x = t,y = 3t + 7, t ∈ R.

In point-normal formula,

~n · (~x− ~p) = 0, ⇒ (3,−1) · (~x− (0, 7)) = 0.

Note that,

(3,−1) · (x, y − 7) = 3x− y + 7 = 0, ⇒ y = 3x + 7.

Therefore, y = 3x + 7 is actually the simplified point-normalformula of the equation of L!

36

Eg 2.8.2: Consider line L:

~x = t~l + ~p = t

[12

]+

[3−2

]=

[t + 3

2t− 2

].

Determine which of the following lines is parallel, perpendic-ular, or neither to L.

(a) ~x =

[2t− 1−t + 3

]; (b) ~x =

[−3t + 6−6t− 5

]; (c) ~x =

[2 + t1− t

].

Ans: We know that ~lL =

[12

].

(a) ~la =

[2−1

], since ~la ·~lL = 2− 2 = 0. La is ⊥ to L.

(b) ~lb =

[−3−6

]= (−3)

[12

]= −3~lL. Lb is ‖ to L.

(c) ~lc =

[1−1

], ~lc ·~lL = 1− 2 = −1.

cos θ =~lc ·~lL||~lc||||~lL||

=−1√

10⇒ θ ≈ 108.435◦.

Thus, Lc is at an angle θ ≈ 108.435◦ to line L.

37

∗2.8.2 Curves in 2D/ Curved surfaces in higher dimensions

(1) Circle of radius R centred at ~c.

||~x− ~c|| = R, (R > 0, R ∈ R.) 0

x =(x,y)− cx

c

x

y

• In R3, it is a sphere of radius R centred at ~c.

• In Rn (n > 3), it is a “hypersphere” of radius R centredat ~c.

(2) Parabola with focus located at ~f = (0, p) and a directrixat y = −p, (p > 0).

||~x− ~f || = y + p, (p > 0, p ∈ R.)

x

x f−

p

x =(x,y)

f y

y

−pdirectrix

In R2 : ||~x−~f ||2 = x2+(y−p)2 = (y+p)2, ⇒ y =1

4px2.

Qusetion: What does ||~x− ~f || = z + p represent in R3?

38

(3) Ellipse with foci located at ~f1 = (−f, 0) and ~f2 = (f, 0),f 2 = a2 − b2.

||~x− ~f1||+||~x− ~f2|| = 2a, (f, a, b > 0.)

12

1=(−f,0)

2=(f,0)

f =a − b2 2 2

1 2f

x =(x,y)f

x f−

x

y

a

b

f−x

f−a

f

−b

In R2, this can be turned into the familiar formx2

a2+y2

b2= 1.

Qusetion: How to derive the equation above?

[ Hint: √(x + f )2 + y2 +

√(x− f )2 + y2 = 2a.

Square both sides, one yields

X +√X2 − 4x2f 2 = 2a2, where X = x2 + y2 + f 2.

Then, square both sides of√X2 − 4x2f 2 = 2a2−X and simplify to yield

a2X = a4 + x2f 2, ⇒ x2(1− f 2

a2) + y2 = a2 − f 2

⇒ x2

a2+y2

b2= 1.

]

39

2.8.3 Planes in 3D and intersections between them

(1) The point-normal formula of a plane in R3:

Let S be a plane in R3. Let ~p be a point on S and ~n be itsnormal.

−

y

Sx

z

S

n

x

y

x

px

z

p

Based on the graph, any point ~x on S must satisfy

~n · (~x− ~p) = 0 or ~n · ~x = ~n · ~p

which is the point-normal formula of a plane in R3. In par-ticular, if ~p = ~0 (i.e. if the plane goes through the origin), theequation reduces to

~n · ~x = 0.

40

Eg 2.8.3: Find the equation of a plane with normal ~n =(1,−1, 1) and a point ~p = (0, 0, 6) on it.

Ans:~n · ~x = ~n · ~p = 6, ⇔ x− y + z = 6.

Eg 2.8.4: For the plane x + 2y − 3z = 3, find its normaland a point on it.

Ans:

x + 2y − 3z = (1, 2,−3) · ~x = ~n · ~x ⇒ ~n = (1, 2,−3).

To find a point, we pick y = z = 0 for simplicity. Thus,x + 2y − 3z = 3 ⇒ x = 3 when y = z = 0. Therefore,~p = (3, 0, 0) is one point on the plane.

Eg 2.8.5: A plane S, x−y+z = 6, and a point ~p0 = (2, 0, 1)are in R3.

(a) Show that ~p0 is not on S.

(b) Find the (shortest) distance between ~p0 and S.

41

Ans:

(a) Plug the point ~p0 = (2, 0, 1) into the equation,

lhs = 2− 0 + 1 = 3 6= 6 = rhs. Thus, ~p0 is not on S.

(b) See the graph, one needs to draw a normal vector that passes through

~p0 which intersects S at ~x0. The distance between ~p0 and ~x0 is the

distance we are looking for.

0

y

x

n

z

S

p0

x

d

Normal of the plane: ~n = (1,−1, 1). The point-direction formula gives

the equation of the normal line that passes through ~p0:

~x = t~n + ~p0 ⇒

xyz

= t

1

−1

1

+

2

0

1

=

t + 2

−tt + 1

.To find the point of intersection between the two, one needs to plug this

into the equation of the plane

x− y + z = (t + 2)− (−t) + (t + 1) = 3t + 3 = 6 ⇒ t∗ = 1.

Thus, ~x0 = ~x(t∗ = 1) = (3,−1, 2). This distance between the two are:

d = ||~x0 − ~p0|| =√

(3− 2)2 + (−1)2 + (2− 1)2 =√

3.

42

(2) The parametric formula of a plane in R3:

Let ~l1, ~l2 be two linearly independent (LI, i.e. the two are notparallel to each other) vectors and ~p be a point on plane S,then the parametric equation of S is given by

~x− ~p = s~l1 + t~l2, (s, t ∈ R are parameters.)

− x

p

l

S

0

1sl

2lt

1

p2x

l

Alternatively,

~x = ~p + s~l1 + t~l2, (s, t ∈ R are parameters.)

43

Eg 2.8.6: A plane S is given in parametric formula as follows.Find its point-normal formula.

~x =

121

+ s

101

+ t

110

.Ans: The normal is

~n = ~l1 ×~l2 =

∣∣∣∣∣∣i j k1 0 11 1 0

∣∣∣∣∣∣ = (−1)i− (−1)j + k = (−1, 1, 1).

When s = t = 0, ~x = ~p = (1, 2, 1) is a point on S. Thus,

~n · ~x = ~n · ~p ⇒ −x + y + z = 2.

Or

x− y − z = −2.

44

Eg 2.8.7: A plane S is given by x + 2y + 3z = 6. Find oneparametric formula of S.

Ans: There are infinitely many apparently different answers!An easier way of doing this will be shown later! For now,One needs to find 3 points on S that are not located on asingle line. For simplicity,

~p0 =

111

, ~p1 =

600

, ~p2 =

002

.Now, let

~l1 = ~p1 − ~p0 =

5−1−1

, ~l2 = ~p2 − ~p0 =

−1−11

.One parametric formula is given by

~x = ~p1 + s~l1 + t~l2 =

600

+ s

5−1−1

+ t

−1−11

.

45

Notice that the method above is not the simplest possible way to find the

parametric form of the equation of a plane in R3.

Because a plane has 2 degrees of freedom, one can assume that z = t and

y = s and substitute into the equation to yield

x + 2s + 3t = 6 ⇒ x = 6− 2s− 3t.

Thus,

~x =

xyz

=

6− 2s− 3t

s

t

=

6

0

0

+ s

−2

1

0

+ t

−3

0

1

.Therefore, given the equation of a plane in point-normal form, e.g. x +

2y+3z = 6, there exist infinitely many ways to express it in the parametric

form. One way to show that the two vectors ~l1 and ~l2 are indeed parallel to

the plane is to show that the cross product between the two is the normal

direction of the plane.

For the solution in previous page,

~l1×~l2 =

∣∣∣∣∣∣i j k

5 −1 −1

−1 −1 1

∣∣∣∣∣∣ = (−2)i− (4)j+(−6)k =

−2

−4

−6

= −2

1

2

3

.For the solution given above,

~l1 ×~l2 =

∣∣∣∣∣∣i j k

−2 1 0

−3 0 1

∣∣∣∣∣∣ = (1)i− (−2)j + (3)k =

1

2

3

.

46

(3) The point-direction (parametric) formula of a line in R3:

Given a point ~p and a the direction ~l of a line L,

~x− ~p = t~l or ~x = t~l + ~p.

l

p

x

L

x

t>0

p

t<0

−

t=0

x

y

z

0

47

(4) The point-normal formula of a line in R3:

Given a point ~p and the normals ~n1, ~n2 of two planes thatintersect at the line,

1

n2

S

S2

1n

L

p

x

xp−

l

o

{~n1 · (~x− ~p) = 0,~n2 · (~x− ~p) = 0.

or

{~n1 · ~x = ~n1 · ~p,~n2 · ~x = ~n2 · ~p.

• Notice that ~l = ~n1 × ~n2.• In parametric formula in any space dimension, the number

of parameters is the dimension of the geometric object: 1parameter for a line, 2 for a plane.

• In non-parametric form inND space, N minus the numberof equations is the dimension of the object: e.g., in 3-D space, 1 equation describes a 3-1=2 D plane, while 2equations describe a 3-2=1 D line.Number of equations=codimension of the object.

48

Eg 2.8.8: A line L in R3 is defined by

{x + y + z = 3,x− y + 2z = −7.

(a) Find the normals of the two planes.

(b) Find the direction of L.

(c) Find a point on L.

(d) Write the equation of L in parametric form.

Ans:

(a) ~n1 = (1, 1, 1), ~n2 = (1,−1, 2).

(b) Since the direction of L, ~l, is ⊥ to ~n1 and ~n2,

~l = ~n1×~n2 =

∣∣∣∣∣∣i j k1 1 11 −1 2

∣∣∣∣∣∣ = (3)i−(1)j+(−2)k = (3,−1,−2).

(c) For simplicity, let z = 0. The equations become{x + y = 3,x− y = −7.

(1)+(2)−→ 2x = −4 −→{x = −2,y = 5.

Thus, ~p = (−2, 5, 0) is one point on L.

(d) ~x = t~l + ~p = t

3−1−2

+

−250

=

3t− 2−t + 5−2t

.49

Eg 2.8.9: A line L in R3 is defined parametrically by

~x = t(1, 2, 1) + (1, 1, 2).

Find equations of two planes that intersect at L.

Ans: There are infinitely many solutions. Starting from

~x =

xyz

=

t + 12t + 1t + 2

,one simply needs to find distinct combinations of x, y, z thatwould cancel all terms that contain t. For example,

x− y + z = (t + 1)− (2t + 1) + (t + 2) = 2.

x + y − 3z = (t + 1) + (2t + 1)− 3(t + 2) = −4.

3x− y − z = 3(t + 1)− (2t + 1)− (t + 2) = 0....

The list can continue to infinity. One can pick any two ofthem, e.g. {

3x− y − z = 0,x− y + z = 2,

are equations of two planes that intersect at line L.

50

2.9 Systems of equations & linear independence

Consider a system of 3 linear, algebraic equations a11x + a12y + a13z = b1,a21x + a22y + a23z = b2, (2.9.1)a31x + a32y + a33z = b3,

where aij, bi, (i, j = 1, 2, 3) are scalars in R.

i jarow # column #

Each eqn defines a plane in R3 with its normal given by

~ni = (ai1, ai2, ai3), (i = 1, 2, 3).

There exist 3 possible situations:

(2 normals in same direction)3 normals are not LI

No common intersection

(on the same plane)

1

Intersect in a line

(not on the same plane)3 normals are LI

Intersect at one point

3 normals are not LI

S

S2

S31S

S2

S3

1S

S2

S3

51

Linear dependence vs linear independence

Note that if ~n1, ~n2 are parallel to each other, then

~n1 = s~n2 or ~n2 = s~n1, (s 6= 0 ∈ R.)

Thus, one is a scalar multiple of the other. Two vectors arelinearly dependent (LD) if they are parallel to each other orif one is a scalar multiple of the other.

On the contrary, if two vectors are not parallel (i.e. havedifferent directions) or cannot be expressed as a scalar multipleof each other, then they are linearly independent (LI).

Geometrically, if the pgram with sides ~n1, ~n2 have non-zero area, then the two are LI. Otherwise, they are LD.

Def: Linear combination (LC). If ~ai, (i = 1, · · · ,m) arevectors in Rn, if

~b = s1~a1 + s2~a2 + · · · + sm~am, (~b 6= ~0)

where the scalars, si (i = 1, · · · ,m) ∈ R, are not simulta-

neously zero, then ~b is called a linear combination (LC) ofvectors ~ai, (i = 1, · · · ,m).

52

Def: Linear dependence (LD) and linear independence (LI).Nonzero vectors ~ai (i = 1, · · · ,m, ~ai 6= ~0) are LI iff

s1~a1 + s2~a2 + · · · + sm~am = ~0

is satisfied only when the scalars si = (i = 1, · · · ,m) aresimultaneously zero, i.e. si = 0 for all i. Otherwise, they areLD. If they are LD, then at least one of them can be expressedas a LC of the others.

Put it in words: A set of vectors are LI if none of themcan be written as a linear combination of others.

Eg: If s1~a1 + s2~a2 + · · · + sm~am = ~0 and s1 6= 0, then

~a1 =

(−s2s1

)~a2 +

(−s3s1

)~a3 + · · · +

(−sms1

)~am,

therefore, ~a1 must be a LC of the other vectors in the set.

Basis of Rn: There are at most n LI vectors in any setof vectors in Rn. Any set of n LI vectors can form a basiswhose linear combinations can generate all vectors in Rn. Inpractice, we often use orthonormal basis, i.e. n unit vectorsthat are mutually orthogonal to each other.

53

Eg 2.9.1: Determine if vectors in the following sets are LI.

(a)

{[1−2

],

[−36

]};

(b)

{[11

],

[−12

]};

(c)

{[11

],

[2−1

],

[2−3

]}.

Ans:

(a) Not, not LI because

[−36

]= (−3)

[1−2

].

Or because Apgram =

∣∣∣∣ 1 −2−3 6

∣∣∣∣ = 6− 6 = 0.

(b) Yes, LI because Apgram =

∣∣∣∣ 1 1−1 2

∣∣∣∣ = 2 + 1 = 3 6= 0. Or,

they are not scalar multiple of each other.

(c) No, they can’t be LI because there could be at most 2 LIvectors in any set of vectors in R2.

54

Eg 2.9.2: Determine if vectors in the following sets are LIor LD.

(a)

0

11

, 0

2−1

, 0

36

;

(b)

1

01

, 1

2−1

;

(c)

1

01

, 1−12

, 1

11

.

Ans:

(a) LD, because all 3 vectors are in the 2D yx-plane.

Or

036

= 5

011

− 0

2−1

.(b) LI, because they are not scalar multiple of each other.

(c) LI, because Vppiped = abs

∣∣∣∣∣∣1 0 11 −1 21 1 1

∣∣∣∣∣∣

= | − 1 + 1 + 0− (−1)− 2− 0| = |1− 2| = 1 6= 0.

55

Summary:

• In R2, 2 vectors are LI if they point to different directions.

• In R3, 3 vectors pointing to 3 different directions can notguarantee that they are LI. They are LI only when theyare NOT in the same plane.

• In Rn, n vectors are LI iff none of them can be expressedas a nontrivial LC of the others. Or when the volumeof the “hyper-parallelepiped” with sides defined by thesevectors is nonzero.

56