Higher Engineering Mathematics

244 VECTOR GEOMETRY

=i(− 12 −84)−j(− 48 −48)

+k(− 168 +24)

=− 96 i+ 96 j− 144 k

or− 48 ( 2 i− 2 j+ 3 k)

Practical applications of vector products

Problem 9. Find the moment and the magni-

tude of the moment of a force of (i+ 2 j− 3 k)

newtons about point B having co-ordinates

(0, 1, 1), when the force acts on a line throughA

whose co-ordinates are (1, 3, 4).

The momentMabout pointBof a force vectorF
which has a position vector ofrfromAis given by:

M=r×F

ris the vector fromBtoA, i.e.r=BA.

ButBA=BO+OA=OA−OB (see Problem 8,

Chapter 21), that is:

r=(i+ 3 j+ 4 k)−(j+k)

=i+ 2 j+ 3 k

Moment,

M=r×F=(i+ 2 j+ 3 k)×(i+ 2 j− 3 k)

=

∣ ∣ ∣ ∣ ∣ ∣

ij k

12 3

12 − 3

∣ ∣ ∣ ∣ ∣ ∣

=i(− 6 −6)−j(− 3 −3)

+k(2−2)

=− 12 i+ 6 jNm

The magnitude ofM,

|M|=|r×F|

=

√

[(r•r)(F•F)−(r•F)^2 ]

r•r=(1)(1)+(2)(2)+(3)(3)= 14

F•F=(1)(1)+(2)(2)+(−3)(−3)= 14

r•F=(1)(1)+(2)(2)+(3)(−3)=− 4

|M|=

√

[14× 14 −(−4)^2 ]

=

√

180 Nm=13.42 Nm

Problem 10. The axis of a circular cylinder

coincides with thez-axis and it rotates with an

angular velocity of (2i− 5 j+ 7 k) rad/s. Deter-

mine the tangential velocity at a pointP on

the cylinder, whose co-ordinates are (j+ 3 k)

metres, and also determine the magnitude of the

tangential velocity.

The velocityvof pointPon a body rotating with

angular velocityωabout a fixed axis is given by:

v=ω×r,

whereris the point on vectorP.

Thusv=(2i− 5 j+ 7 k)×(j+ 3 k)

=

∣ ∣ ∣ ∣ ∣ ∣

ijk

2 − 57

013

∣ ∣ ∣ ∣ ∣ ∣

=i(− 15 −7)−j(6−0)+k(2−0)

=(− 22 i− 6 j+ 2 k)m/s

The magnitude ofv,

|v|=

√

[(ω•ω)(r•r)−(r•ω)^2 ]

ω•ω=(2)(2)+(−5)(−5)+(7)(7)= 78

r•r=(0)(0)+(1)(1)+(3)(3)= 10

ω•r=(2)(0)+(−5)(1)+(7)(3)= 16

Hence,

|v|=

√

(78× 10 − 162 )

=

√

524 m/s=22.89 m/s

Now try the following exercise.

Exercise 98 Further problems on vector

products

In problems 1 to 4, determine the quantities

stated when

p= 3 i+ 2 k,q=i− 2 j+ 3 kand

r=− 4 i+ 3 j−k

1. (a)p×q (b)q×p
   [(a) 4i− 7 j− 6 k(b)− 4 i+ 7 j+ 6 k]


2. (a)|p×r| (b)|r×q|
   [(a) 11.92 (b) 13.96]