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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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256 Chapter 7 Vectors

16 Show that the vectorsi −j and −3i −3j are

perpendicular.

17 Find the norm ofeach ofthe vectors

⎛ ⎞ ⎛ ⎞

7 2

⎜ 2

⎝ −1⎠ and ⎜ 1

⎝ 0 ⎠

2 −4

18 (a)Usethe scalar product to findthe value ofthe

scalar µ sothat i +j+µk isperpendicular to the

vector i +j+k.

(b)Usethevectorproductandtheresultsfrompart(a)

to findamutually perpendicular setofunitvectors

ˆv 1 , ˆv 2 and ˆv 3 , where ˆv 1 is inclined equally to the

vectorsi,jand k.

19 ThepointsA, Band C have coordinates (2,−1, −2),

(4,−1,−3)and (1,3,−1).

(a) Writedown the vectors → ABand → AC.

(b) Using the vectorproduct findaunitvector

which isperpendicular to the plane containing

A, B and C.

(c) IfDisthe point with coordinates (3,0,1),use

the scalar product to find the perpendicular

distance fromDto the plane ABC.

20 Thecondition forvectorsa,band cto be coplanar

(i.e.they liein the same plane) isa · (b ×c) = 0.

(a) Show that the vectorsa = 4i +5j +6k,

b=6i−3j−3kandc=−i+2j+2kare

notcoplanar.

(b) Givend = −i +2j + λk,findthe value of λ so

thata, b anddare coplanar.

21 Points Aand Bhave position vectorsaand b

respectively. Show that the position vector ofan

arbitrarypoint on the line AB isgiven by

r = a + λ (b −a) forsome scalar λ.Thisis the

vector equationofthe line.

22 Usevector methods to show that the threemedians of

any triangle intersect atacommon point (called the

centroid).

23 Usethe vectorproduct to findthe area ofatriangle

with vertices atthe pointswith coordinates (1,2,3),

(4,−3,2),and (8,1,1).

Solutions

1 (a) 24, −7i −32j +17k (b) 18, 0

2 0

3 √ 54, √ 66,57,0.9548

4 √ 41, √ 19, √ 154,0.4446

5

1

139

a,

1

20

b,

6 5i+7j−4k,22

1

2296

(−12i −6j −46k)

7 (a×b)×c=4i−12j−16k

a·c=8,b·c=4

a×(b×c)=−2i−22j−8k

1 1

9 √ (i +7j), √ (i +2j),18.4 ◦

50 5

10

µq

56 √ (i +11j +7k)

14π

12 (a) 3i−5j−4k

(b) 5j −2k

13 √ 24 = 4.90

14 87.5 ◦

15

1

230

(5i +14j −3k)

17 √ 58, √ 21

18 (a) µ=−2

(b) ˆv 1 = 1 √

3

(1,1,1), ˆv 2 = 1 √

6

(1,1,−2),

ˆv 3 = 1 √

2

(−1,1,0)

19 (a) (2,0,−1), (−1,4,1) (b)

1

(4,−1,8) (c) 3

9

20 (a) a · (b ×c) = 9 ≠ 0 andhence the vectors are not

coplanar

(b) λ = 31/14

23 1 2√

1106

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