Begin2.DVI

I6-4.
By construction

A~+^1

2

B~=C,~ B~+^1

2

A~=D,~ A~+E~=B~

All these vectors are coplaner so that there exists scalar

constantsα,β,γ,δsuch that

A~+αE~=βC~ and A~+γE~=δD~

or

A~+α(B~−A~) =β(A~+^1

2

B~) and A~+γ(B~−A~) =δ(B~+^1

2

A~)

This implies that

A~(1−α−β) +B~(α−^1

2

β) =~ 0 and A~(1−γ−^1

2

δ) +B~(γ−δ) =~ 0

This produces the simultaneous equations

α+β= 1, α−

1

2 β= 0, γ+

1

2 δ= 1, γ−δ= 0

Solving forα,β,γ,δone finds

α= 1/ 3 , β= 2/ 3 , γ= 2/ 3 , δ= 2/ 3

I6-5. (a)c 1 A~+c 2 B~+c 3 C~=~ 0 gives the system of equations

c 1 − 4 c 2 + 7c 3 =0

c 1 − 3 c 2 + 6c 3 =0

− 2 c 1 − 6 c 3 =0

=⇒ c 1 =− 3 c 2 , c 3 =c 2

Sincec 26 = 0, selectc 2 = 1for convenience, thenc 1 =− 3 , c 2 = 1, c 3 = 1, so vectors are

linearly dependent.

(b) Linearly independent

(c) Linearly independent

I6-6. The vectorsA,~ B,~ C~are linearly dependent if and only ifA~·(B~×C~) = 0

IfA~·(B~×C~) = 0andA~ 6 = 0, thenB~×C~=~ 0 which implies the vectorsB~ andC~ are

colinear. If the determinant

∣∣

∣∣

∣∣

A 1 A 2 A 3

B 1 B 2 B 3

C 1 C 2 C 3

∣∣

∣∣

∣∣= 0, then two rows of the determinant

are proportional which implies two of the vectors are colinear.

Solutions Chapter 6