282 Higher Engineering Mathematics
Hence
p × (2r × 3q) = (4i + j − 2k)
× (−24i − 42j − 12k)
=
i
j
k
4
1 −2
−24 −42 −12
= i(−12 − 84) − j(−48 − 48)
+ k(−168 + 24)
= −96i +96j − 144k
or −48(2i − 2j +3k)
Practical applications of vector products
Problem 9. Find the moment and the magnitude
of the moment of a force of (i + 2j −3k) newtons
about point B having co-ordinates (0, 1, 1), when
the force acts on a line through A whose
co-ordinates are (1, 3, 4).
The moment M about point B of a force vector F which
has a position vector of r from A is given by:
M = r × F
r is the vector from B to A, i.e. r = BA.
But BA = BO + OA = OA − OB (see Problem 13,
page 262), that is:
r = (i + 3j + 4k) − ( j + k)
= i + 2j + 3k
Moment,
M = r × F = (i + 2j + 3k) × (i + 2j − 3k)
=
i j
k
1 2
3
1 2 −3
= i(−6 − 6) − j(−3 − 3)
+ k(2 − 2)
= −12i + 6j Nm
The magnitude of M,
|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 the z-axis and it rotates with an
angular velocity of (2i − 5j + 7k) rad/s. Determine
the tangential velocity at a point P on the cylinder,
whose co-ordinates are ( j + 3k) metres, and also
determine the magnitude of the tangential velocity.
The velocity v of point P on a body rotating with angular
velocity ω about a fixed axis is given by:
v = ω × r,
where r is the point on vector P.
Thus
v = (2i − 5j + 7k) × ( j + 3k)
=
i
j k
2 −5 7
0
1 3
= i(−15 − 7) − j(6 − 0) + k(2 − 0)
= (−22i − 6j +2k) m/s
The magnitude of v,
|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 − 16 2 )
=
√
524 m/s = 22.89 m/s
Hence
p × (2r × 3q) = (4i + j − 2k)
× (−24i − 42j − 12k)
=
i
j
k
4
1 −2
−24 −42 −12
= i(−12 − 84) − j(−48 − 48)
+ k(−168 + 24)
= −96i +96j − 144k
or −48(2i − 2j +3k)
Practical applications of vector products
Problem 9. Find the moment and the magnitude
of the moment of a force of (i + 2j −3k) newtons
about point B having co-ordinates (0, 1, 1), when
the force acts on a line through A whose
co-ordinates are (1, 3, 4).
The moment M about point B of a force vector F which
has a position vector of r from A is given by:
M = r × F
r is the vector from B to A, i.e. r = BA.
But BA = BO + OA = OA − OB (see Problem 13,
page 262), that is:
r = (i + 3j + 4k) − ( j + k)
= i + 2j + 3k
Moment,
M = r × F = (i + 2j + 3k) × (i + 2j − 3k)
=
i j
k
1 2
3
1 2 −3
= i(−6 − 6) − j(−3 − 3)
+ k(2 − 2)
= −12i + 6j Nm
The magnitude of M,
|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 the z-axis and it rotates with an
angular velocity of (2i − 5j + 7k) rad/s. Determine
the tangential velocity at a point P on the cylinder,
whose co-ordinates are ( j + 3k) metres, and also
determine the magnitude of the tangential velocity.
The velocity v of point P on a body rotating with angular
velocity ω about a fixed axis is given by:
v = ω × r,
where r is the point on vector P.
Thus
v = (2i − 5j + 7k) × ( j + 3k)
=
i
j k
2 −5 7
0
1 3
= i(−15 − 7) − j(6 − 0) + k(2 − 0)
= (−22i − 6j +2k) m/s
The magnitude of v,
|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 − 16 2 )
=
√
524 m/s = 22.89 m/s