Higher Engineering Mathematics

SCALAR AND VECTOR PRODUCTS 245

D

3. (a) 2p× 3 r(b) (p+r)×q
   [
   (a)− 36 i− 30 j− 54 k
   (b) 11i+ 4 j−k

]

4. (a)p×(r×q) (b) (3p× 2 r)×q
   [
   (a)− 22 i−j+ 33 k
   (b) 18i+ 162 j+ 102 k

]

5. For vectorsp= 4 i−j+ 2 kand
   q=− 2 i+ 3 j− 2 kdetermine: (i)p•q
   (ii)p×q(iii)|p×q|(iv)q×pand (v) the
   angle between the vectors.
   ⎡

⎣

(i)−15 (ii)− 4 i+ 4 j+ 10 k

(iii) 11.49 (iv) 4i− 4 j− 10 k

(v) 142. 55 ◦

⎤

⎦

6. For vectorsa=− 7 i+ 4 j+^12 kandb= 6 i−
   5 j−k find (i) a•b (ii) a×b (iii)|a×b|
   (iv) b×aand (v) the angle between the
   vectors. ⎡

⎢

⎣

(i)− 6212 (ii)− 112 i− 4 j+ 11 k

(iii) 11.80 (iv) 1^12 i+ 4 j− 11 k

(v) 169. 31 ◦

⎤

⎥

⎦

7. Forces of (i+ 3 j), (− 2 i−j), (i− 2 j) newtons
   act at three points having position vectors of
   (2i+ 5 j), 4jand (−i+j) metres respectively.
   Calculate the magnitude of the moment.
   [10 Nm]


8. A force of (2i−j+k) newtons acts on a line
   through pointPhaving co-ordinates (0, 3, 1)
   metres. Determine the moment vector and its
   magnitude about pointQhaving co-ordinates
   (4, 0,−1) metres.
   [
   M=(5i+ 8 j− 2 k)Nm,
   |M|= 9 .64 Nm

]

9. A sphere is rotating with angular velocityω
   about thez-axis of a system, the axis coincid-
   ing with the axis of the sphere. Determine the
   velocity vector and its magnitude at position
   (− 5 i+ 2 j− 7 k) m, when the angular velocity
   is (i+ 2 j) rad/s. [
   υ=− 14 i+ 7 j+ 12 k,
   |υ|= 19 .72 m/s

]

10. Calculate the velocity vector and its magni-
    tude for a particle rotating about thez-axis
    at an angular velocity of (3i−j+ 2 k) rad/s
    when the position vector of the particle is at
    (i− 5 j+ 4 k)m.
    [6i− 10 j− 14 k, 18.22 m/s]

22.4 Vector equation of a line

The equation of a straight line may be determined,

given that it passes through the pointAwith position

vectorarelative toO, and is parallel to vectorb. Let

rbe the position vector of a pointPon the line, as

shown in Fig. 22.10.

b

a

r

A

P

O

Figure 22.10

By vector addition,OP=OA+AP,

i.e.r=a+AP.

However, as the straight line throughAis parallel

to the free vectorb(free vectormeans one that

has the same magnitude, direction and sense), then

AP=λb, whereλis a scalar quantity. Hence, from

above,

r=a+λb (8)

If, say,r=xi+yj+zk,a=a 1 i+a 2 j+a 3 kand

b=b 1 i+b 2 j+b 3 k, then from equation (8),

xi+yj+zk=(a 1 i+a 2 j+a 3 k)

+λ(b 1 i+b 2 j+b 3 k)

Hencex=a 1 +λb 1 ,y=a 2 +λb 2 andz=a 3 +λb 3.

Solving forλgives:

x−a 1

b 1

=

y−a 2

b 2

=

z−a 3

b 3

=λ (9)

Equation (9) is the standard Cartesian form for the

vector equation of a straight line.

Problem 11. (a) Determine the vector equa-

tion of the line through the point with position

vector 2i+ 3 j−kwhich is parallel to the vector

i− 2 j+ 3 k.