CHAPTER 51 SCALAR AND VECTOR PRODUCTS
CHAPTER 51 SCALAR AND VECTOR PRODUCTS EXERCISE 216 Page 591 1. Find the scalar product a • b when (i) a = i + 2j – k and b = 2i + 3j + k (ii) a = i – 3j + k and b = 2i + j + k
(a) a b = (1)(2) + (2)(3) + (–1)(1) = 7 (b) a b = (1)(2) + (–3)(1) + (1)(1) = 2 – 3 + 1 = 0
2. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p • q
(b) p • r
(a) p q = (2)(0) + (–3)(4) + (0)(–1) = –12 (b) p r = (2)(1) + (–3)(2) + (0)(–3) = 2 – 6 = –4
3. Given q = 4j – k and r = i + 2j – 3k, determine: (a) q • r (b) r • q
(a) q r = (0)(1) + (4)(2) + (–1)(–3) = 0 + 8 + 3 = 11 (b) r q = (1)(0) + (2)(4) + (–3)(–1) = 0 + 8 + 3 = 11
4. Given p = 2i – 3j and r = i + 2j – 3k, determine: (a) p (b) r
(a) p =
( (2)
(b) r =
( (1)
)
2
+ (−3) 2 + (0) 2 = 13
2
+ (2) 2 + (−3) 2 = 14
)
5. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p •(q + r)
(b) 2r •(q – 2p)
(a) p ⋅ ( q + r ) = (2i – 3j)(4j – k + i + 2j – 3k) = (2i – 3j)( i + 6j – 4k) = (2)(1) + (–3)(6) + (0)(–4) = 2 – 18 + 0 = –16 (b) 2r ⋅ (q – 2p) = (2i + 4j – 6k) ⋅ (4j – k – 2(2i – 3j)) = (2i + 4j – 6k) ⋅ (– 4i + 10j – k) 886
© 2014, John Bird
= (2)(–4) + (4)(10) + (–6)(–1) = – 8 + 40 + 6 = 38
6. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p + r
( (2 + 1)
(a) p + r =
( (2)
(b) p + r =
2
2
(b) p + r
)
+ (−3 + 2) 2 + (0 − 3) 2 = 19
)
+ (−3) 2 + (0) 2 +
((1) + ( 2) + ( −3) ) = 2
2
2
13 + 14 = 7.347
7. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, find the angle between (a) p and q (b) q and r
(a) From equation (4), page 589, cos θ =
=
a + a2 2 + a32 b12 + b2 2 + b32 (2)(0) + (−3)(4) + (0)(−1) 2 + (−3) + 0 2
2
2
0 + 4 + (−1) 2
2
2
=
−12 = –0.8072 13 17
θ = cos −1 (−0.8072) = 143.82°
from which, (b) cos θ =
a1b1 + a2b2 + a3b3 2 1
(0)(1) + (4)(2) + (−1)(−3) = 0 + 42 + (−1) 2 12 + 22 + (−3) 2 2
11 = 0.713024 17 14
θ = cos −1 (0.713024) = 44.52°
from which,
8. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the direction cosines of: (a) p
(b) q
(c) r
(a) For p, cos α =
2 = 2 2 + (−3) 2
0 −3 2 = 0.555, cos β = = –0.832 and cos γ = =0 13 13 13
(b) For q, cos α =
0 = 2 4 + (−1) 2
−1 4 0 = 0, cos β = = 0.970 and cos γ = = –0.243 17 17 17
1 = 2 2 1 + 2 + (−3) 2 −3 and cos γ = = –0.802 14
(c) For r, cos α =
2 1 = 0.267, cos β = = 0.535 14 14
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© 2014, John Bird
9. Determine the angle between the forces: F1 = 3i + 4j + 5k and F2 = i + j + k
cos θ =
(3)(1) + (4)(1) + (5)(1) = 32 + 42 + 52 12 + 12 + 12
from which,
12 = 0.979795... 50 3
θ = cos −1 (0.979795...) = 11.54°
10. Find the angle between the velocity vectors v 1 = 5i + 2j + 7k and v2 = 4i + j – k
cos θ =
(5)(4) + (2)(1) + (7)(−1) = 52 + 22 + 7 2 42 + 12 + (−1) 2
from which,
15 = 0.40032 78 18
θ = cos −1 (0.40032) = 66.40°
11. Calculate the work done by a force F = (–5i + j + 7k) N when its point of application moves from point (–2i – 6j + k) m to the point (i – j + 10k) m. Work done = F ⋅ d where d = (i – j + 10k) – (–2i – 6j + k) = 3i + 5j + 9k Hence, work done = (–5i + j + 7k) ⋅ (3i + 5j + 9k) = –15 + 5 + 63 = 53 N m
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© 2014, John Bird
EXERCISE 217 Page 594
1. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × q
(b) q × p
i j k 0 2 3 2 3 0 (a) p × q = 3 0 2 = i –j +k = 4i – 7j – 6k 1 3 1 −2 −2 3 1 −2 3 i j k −2 3 1 3 1 −2 (b) q × p = 1 −2 3 = i –j +k = –4i + 7j + 6k 0 2 3 2 3 0 3 0 2
2. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × r
(b) r × q
( p ⋅ p )( r ⋅ r ) − ( p ⋅ r )
(a) p × r =
where p ⋅ p = (3)(3) + (2)(2) = 13,
2
r ⋅ r = (–4)(–4) + (3)(3) + (–1)(–1) = 26 and Hence,
p×r =
(b) r × q = =
2 (13)(26) − (−14) =
( r ⋅ r )( q ⋅ q ) − ( r ⋅ q )
2
p ⋅ r = (3)(–4) + (0)(3) + (2)(–1) = –14 (338 − 196) =
142 = 11.92
where r ⋅ r = (–4)(–4) + (3)(3) + (–1)(–1) = 26, q ⋅ q = (1)(1) +(–2)(–2) + (3)(3) = 14
and Hence,
r×q =
2 (26)(14) − (−13) =
r ⋅ q = (–4)(1) + (3)(–2) + (–1)(3) = –13 (364 − 169) =
195 = 13.96
3. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) 2p × 3r
(b) (p + r) × q
(a) 2p = 2(3i + 2k) = 6i + 4k
and 3r = 3(– 4i + 3j – k) = – 12i + 9j – 3k
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© 2014, John Bird
i j k 0 Hence, 2p × 3r = 6 0 4 =i 9 − 12 9 −3
4 6 –j −3 − 12
4 6 0 +k = – 36i – 30j + 54k −3 − 12 9
(b) p + r = 3i + 2k + (– 4i + 3j – k) = – i + 3j + k i Hence, (p + r) × q = − 1
j 3
1 −2
k 1 =i 3
−1 1 3 1 −1 –j +k −2 3 1 3 1
3 = 11i + 4j – k −2
4. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × (r × q)
(b) (3p × 2r) × q
i j k (a) r × q = − 4 3 −1 = 7i + 11j + 5k 1 −2 3
Hence,
i j k p × (r × q) = 3 0 2 = (–22)i – (1)j + (33)k = –22i – j + 33k 7 11 5
(b) 3p × 2r =
i 9
j 0
k 6
= (–36)i – (–18 + 48)j + (54)k = –36i – 30j + 54k
− 8 6 −2 i j k (3p × 2r) × q = − 36 −30 54 = (–90 + 108)i – (–108 – 54)j + (72 + 30)k 1
−2
3
= 18i + 162j + 102k
5. For vectors p = 4i – j + 2k and q = –2i + 3j – 2k determine (i) p • q
(ii) p × q
(iii) p × q
(iv) q × p and (v) the angle between the vectors
(i)
p • q = (4)(– 2) + (–1)(3) +(2)(–2) = –15
(ii)
i j k p × q = 4 −1 2 = i ( 2 − 6 ) – j(– 8 + 4) + k(12 – 2) = −4i + 4 j + 10k − 2 3 −2
(iii) p ⋅ p = (4)(4) + (–1)(–1) + ( 2 )( 2 ) = 21 890
© 2014, John Bird
q ⋅ q = (–2)(–2) + (3)(3) + (–2)(–2) = 17 p ⋅ q = (4)(–2) + (–1)(3) + ( 2 )( −2 ) = –15 p×q =
= ( 357 − 225 )
2 ( 21)(17 ) − ( −15= )
i
132 = 11.49
k (iv) q × p = − 2 3 −2 = i ( 6 − 2 ) – j(– 4 + 8) + k(2 – 12) = 4i − 4 j − 10k 4 −1 2
(v)
cos θ =
j
(4)(−2) + (−1)(3) + ( 2 ) (−2) (4) 2 + (−1) 2 + ( 2 )
6. For vectors a = –7i + 4j +
(i)
(−2) 2 + (3) 2 + (−2) 2
=
−15 = −0.793884... 21 17
θ = cos −1 (−0.793884...) = 142.55°
from which,
(i) a • b
2
(ii) a × b
1 k and b = 6i – 5j – k find 2
(iii) a × b
(iv) b × a and (v) the angle between the vectors
1 1 a ⋅ b = (–7)(6) + (4)(–5) + ( )(–1) = –62 2 2
i (ii) a × b = − 7
6
j
k
5 1 1 = i −4 + – j(7 – 3) + k(35 – 24) = −1 i − 4 j + 11k 2 2 2 −5 −1 4
1 1 (iii) a ⋅ a = (–7)(–7) + (4)(4) + = 65.25 2 2 b ⋅ b = (6)(6) + (–5)(–5) + (–1)(–1) = 62
1 a ⋅ b = (–7)(6) + (4)(–5) + ( −1) = –62.5 2 a×b =
2 ( 65.25 )( 62 ) − ( −62.5 = )
i
(iv) b × a =
j
= ( 4045.5 − 3906.25 )
139.25 = 11.80
k
1 5 −5 −1 = i − + 4 – j(3 – 7) + k(24 – 35) = 1 i + 4 j − 11k 2 2 1 −7 4 2
6
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© 2014, John Bird
cos θ =
(v)
1 (−7)(6) + (4)(−5) + (−1) 2 1 (−7) 2 + (4) 2 + 2
2
=
(6) 2 + (−5) 2 + (−1) 2
−62.5 = −0.9826388 65.25 62
θ = cos −1 (−0.9826388) = 169.31°
from which,
7. Forces of (i + 3j), (– 2i – j), (i – 2j) newtons act at three points having position vectors of (2i + 5j), 4j and (–i + j) metres, respectively. Calculate the magnitude of the moment.
Magnitude of the moments = the sum of the magnitudes of the three forces When position vector, r = 2i + 5j and F = i + 3j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (2)(2) + (5)(5) +(0)(0) = 29 F ⋅ F = (1)(1) + (3)(3) +(0)(0) = 10 r ⋅ F = (2)(1) + (5)(3) + (0)(0) = 17
i.e.
M =
( 29 )(10 ) − (17 )
2
=1Nm
When position vector, r = 4j and F = –2i – j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (0)(0) + (4)(4) +(0)(0) = 16 F ⋅ F = (– 2)(–2) + (–1)(–1) +(0)(0) = 5 r ⋅ F = (0)(–2) + (4)(–1) + (0)(0) = –4
i.e.
M =
(16 )( 5) − ( −4 )
2
=8Nm
When position vector, r = –i + j and F = –i – 2j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (1)(1) + (1)(1) + (0)(0) = 2 F ⋅ F = (–1)(–1) + (–2)(–2) + (0)(0) = 5 892
© 2014, John Bird
r ⋅ F = (1)(–1) + (1)(–2) + (0)(0) = –3 i.e.
M =
( 2 )( 5) − ( −3)
2
=1Nm
Hence, magnitude of the moment = 1 + 8 + 1 = 10 N m
8. A force of (2i – j + k) newtons acts on a line through point P having coordinates (0, 3, 1) metres. Determine the moment vector and its magnitude about point Q having coordinates (4, 0, –1) metres.
Position vector, r = (0i + 3j + k) – (4i + 0j – k) = –4i + 3j + 2k
Moment, M = r × F
i j k where M = − 4 3 2 = (3 + 2)i – (–4 – 4)j + (4 – 6)k 2 −1 1 = (5i + 8j – 2k) N m
Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (–4)(–4) + (3)(3) + (2)(2) = 29 F ⋅ F = (2)(2) + (–1)(–1) +(1)(1) = 6 r ⋅ F = (–4)(2) + (3)(–1) +(2)(1) = –9
i.e.
M =
( 29 )( 6 ) − ( −9 )
2
= 9.64 N m
9. A sphere is rotating with angular velocity ω about the z-axis of a system, the axis coinciding with the axis of the sphere. Determine the velocity vector and its magnitude at position (–5i + 2j – 7k) m, when the angular velocity is (i + 2j) rad/s.
Velocity vector, v = ω × r = (i + 2j ) × (–5i + 2j – 7k) =
i
j
k
1
2
0
− 5 2 −7
= (–14 – 0)i – (– 7 – 0)j + (2 + 10)k = –14i + 7j + 12k
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Magnitude of v, v =
(ω ⋅ ω )( r ⋅ r ) − ( r ⋅ ω )
2
where ω ⋅ ω = (1)(1) + (2)(2) + (0)(0) = 5 r ⋅ r = (– 5)(–5) + (2)(2) + (–7)(–7) = 78 r ⋅ ω = (–5)(1) + (2)(2) + (–7)(0) = –1 Hence,
v =
( 5)( 78) − ( −1)
2
= 19.72 m/s
10. Calculate the velocity vector and its magnitude for a particle rotating about the z-axis at an angular velocity of (3i – j + 2k) rad/s when the position vector of the particle is at (i – 5j + 4k) m.
i j k Velocity vector, v = ω × r = (3i – j + 2k) × (i – 5j + 4k) = 3 1 2 1 −5 4
= (–4 + 10)i – (12 – 2)j + (–15 + 1)k = 6i – 10j – 14k Magnitude of v, v =
(ω ⋅ ω )( r ⋅ r ) − ( r ⋅ ω )
2
where ω ⋅ ω = (3)(3) + (–1)(–1) + (2)(2) = 14 r ⋅ r = (1)(1) + (–5)(–5) + (4)(4) = 42 r ⋅ ω = (1)(3) + (–5)(–1) + (4)(2) = 16 Hence,
v =
(14 )( 42 ) − (16 )
2
= 18.22 m/s
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© 2014, John Bird
EXERCISE 218 Page 596
1. Find the vector equation of the line through the point with position vector 5i – 2j + 3k which is parallel to the vector 2i + 7j – 4k. Determine the point on the line corresponding to λ = 2 in the resulting equation. Vector equation of the line, r = a + λb = (5i – 2j + 3k) + λ(2i + 7j – 4k) i.e.
r = (5 + 2λ)i + (7λ – 2)j + (3 – 4λ)k
When λ = 2,
r = 9i + 12j – 5k
2. Express the vector equation of the line in Problem 1 in standard Cartesian form.
The vector equation of a straight line in standard Cartesian form is:
x − a1 y − a2 z − a3 = = = λ b1 b2 b3 Since a = 5i – 2j + 3k,
a1 = 5, a2 = −2 and a3 = 3
and
b1 = 2, b2 = 7 and b3 = −4
b = 2i + 7j – 4k,
then the Cartesian equations are:
3. Express in vector form:
3x − 1 5 y + 1 4 − z = = 4 2 3
i.e.
i.e.
x −5 y + 2 z −3 = = = λ 2 7 −4
or
x −5 y + 2 3− z = = = λ 2 7 4
3x − 1 5 y + 1 4 − z = = 3 4 2
i.e.
1 1 3 x − 5 y + 3 5 −1( z − 4 ) = = 4 2 3
1 1 x− y+ 3 5 = = 4 2 3 5
( z − 4) −3
1 1 a1 = , a2 = − and a3 = 4 3 5 895
© 2014, John Bird
and
b1 =
4 2 , b2 = and b3 = −3 3 5
Hence, in vector form the equation is: r = ( a1 + λb1 ) i + ( a2 + λb2 ) j + ( a3 + λb3 ) k
1 4 = + λi + 3 3 or
r=
1 2 − + λ j + ( 4 − 3λ ) k 5 5
1 1 (1 + 4λ ) i + ( 2λ − 1) j + ( 4 − 3λ ) k 3 5
4. Express in vector form: 2x + 1 =
2 x + 1 1 − 4 y 3z − 1 = = 1 5 4
i.e.
3z − 1 1− 4 y = 4 5
1 1 1 z− x+ y− 3 2 = 4 = 1 5 4 − 2 4 3
i.e.
1 1 1 and a3 = a1 = − , a2 = 2 4 3
and
1 5 4 b1 = , b2 = − and b3 = 2 4 3
Hence, in vector form the equation is: r = ( a1 + λb1 ) i + ( a2 + λb2 ) j + ( a3 + λb3 ) k
1 5 1 1 1 4 = − + λi + − λj + + λk 4 4 2 2 3 3 or
r=
1 1 1 ( λ − 1) i + (1 − 5λ ) j + (1 + 4λ ) k 2 4 3
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© 2014, John Bird
(a) a b = (1)(2) + (2)(3) + (–1)(1) = 7 (b) a b = (1)(2) + (–3)(1) + (1)(1) = 2 – 3 + 1 = 0
2. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p • q
(b) p • r
(a) p q = (2)(0) + (–3)(4) + (0)(–1) = –12 (b) p r = (2)(1) + (–3)(2) + (0)(–3) = 2 – 6 = –4
3. Given q = 4j – k and r = i + 2j – 3k, determine: (a) q • r (b) r • q
(a) q r = (0)(1) + (4)(2) + (–1)(–3) = 0 + 8 + 3 = 11 (b) r q = (1)(0) + (2)(4) + (–3)(–1) = 0 + 8 + 3 = 11
4. Given p = 2i – 3j and r = i + 2j – 3k, determine: (a) p (b) r
(a) p =
( (2)
(b) r =
( (1)
)
2
+ (−3) 2 + (0) 2 = 13
2
+ (2) 2 + (−3) 2 = 14
)
5. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p •(q + r)
(b) 2r •(q – 2p)
(a) p ⋅ ( q + r ) = (2i – 3j)(4j – k + i + 2j – 3k) = (2i – 3j)( i + 6j – 4k) = (2)(1) + (–3)(6) + (0)(–4) = 2 – 18 + 0 = –16 (b) 2r ⋅ (q – 2p) = (2i + 4j – 6k) ⋅ (4j – k – 2(2i – 3j)) = (2i + 4j – 6k) ⋅ (– 4i + 10j – k) 886
© 2014, John Bird
= (2)(–4) + (4)(10) + (–6)(–1) = – 8 + 40 + 6 = 38
6. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine: (a) p + r
( (2 + 1)
(a) p + r =
( (2)
(b) p + r =
2
2
(b) p + r
)
+ (−3 + 2) 2 + (0 − 3) 2 = 19
)
+ (−3) 2 + (0) 2 +
((1) + ( 2) + ( −3) ) = 2
2
2
13 + 14 = 7.347
7. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, find the angle between (a) p and q (b) q and r
(a) From equation (4), page 589, cos θ =
=
a + a2 2 + a32 b12 + b2 2 + b32 (2)(0) + (−3)(4) + (0)(−1) 2 + (−3) + 0 2
2
2
0 + 4 + (−1) 2
2
2
=
−12 = –0.8072 13 17
θ = cos −1 (−0.8072) = 143.82°
from which, (b) cos θ =
a1b1 + a2b2 + a3b3 2 1
(0)(1) + (4)(2) + (−1)(−3) = 0 + 42 + (−1) 2 12 + 22 + (−3) 2 2
11 = 0.713024 17 14
θ = cos −1 (0.713024) = 44.52°
from which,
8. Given p = 2i – 3j, q = 4j – k and r = i + 2j – 3k, determine the direction cosines of: (a) p
(b) q
(c) r
(a) For p, cos α =
2 = 2 2 + (−3) 2
0 −3 2 = 0.555, cos β = = –0.832 and cos γ = =0 13 13 13
(b) For q, cos α =
0 = 2 4 + (−1) 2
−1 4 0 = 0, cos β = = 0.970 and cos γ = = –0.243 17 17 17
1 = 2 2 1 + 2 + (−3) 2 −3 and cos γ = = –0.802 14
(c) For r, cos α =
2 1 = 0.267, cos β = = 0.535 14 14
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© 2014, John Bird
9. Determine the angle between the forces: F1 = 3i + 4j + 5k and F2 = i + j + k
cos θ =
(3)(1) + (4)(1) + (5)(1) = 32 + 42 + 52 12 + 12 + 12
from which,
12 = 0.979795... 50 3
θ = cos −1 (0.979795...) = 11.54°
10. Find the angle between the velocity vectors v 1 = 5i + 2j + 7k and v2 = 4i + j – k
cos θ =
(5)(4) + (2)(1) + (7)(−1) = 52 + 22 + 7 2 42 + 12 + (−1) 2
from which,
15 = 0.40032 78 18
θ = cos −1 (0.40032) = 66.40°
11. Calculate the work done by a force F = (–5i + j + 7k) N when its point of application moves from point (–2i – 6j + k) m to the point (i – j + 10k) m. Work done = F ⋅ d where d = (i – j + 10k) – (–2i – 6j + k) = 3i + 5j + 9k Hence, work done = (–5i + j + 7k) ⋅ (3i + 5j + 9k) = –15 + 5 + 63 = 53 N m
888
© 2014, John Bird
EXERCISE 217 Page 594
1. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × q
(b) q × p
i j k 0 2 3 2 3 0 (a) p × q = 3 0 2 = i –j +k = 4i – 7j – 6k 1 3 1 −2 −2 3 1 −2 3 i j k −2 3 1 3 1 −2 (b) q × p = 1 −2 3 = i –j +k = –4i + 7j + 6k 0 2 3 2 3 0 3 0 2
2. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × r
(b) r × q
( p ⋅ p )( r ⋅ r ) − ( p ⋅ r )
(a) p × r =
where p ⋅ p = (3)(3) + (2)(2) = 13,
2
r ⋅ r = (–4)(–4) + (3)(3) + (–1)(–1) = 26 and Hence,
p×r =
(b) r × q = =
2 (13)(26) − (−14) =
( r ⋅ r )( q ⋅ q ) − ( r ⋅ q )
2
p ⋅ r = (3)(–4) + (0)(3) + (2)(–1) = –14 (338 − 196) =
142 = 11.92
where r ⋅ r = (–4)(–4) + (3)(3) + (–1)(–1) = 26, q ⋅ q = (1)(1) +(–2)(–2) + (3)(3) = 14
and Hence,
r×q =
2 (26)(14) − (−13) =
r ⋅ q = (–4)(1) + (3)(–2) + (–1)(3) = –13 (364 − 169) =
195 = 13.96
3. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) 2p × 3r
(b) (p + r) × q
(a) 2p = 2(3i + 2k) = 6i + 4k
and 3r = 3(– 4i + 3j – k) = – 12i + 9j – 3k
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i j k 0 Hence, 2p × 3r = 6 0 4 =i 9 − 12 9 −3
4 6 –j −3 − 12
4 6 0 +k = – 36i – 30j + 54k −3 − 12 9
(b) p + r = 3i + 2k + (– 4i + 3j – k) = – i + 3j + k i Hence, (p + r) × q = − 1
j 3
1 −2
k 1 =i 3
−1 1 3 1 −1 –j +k −2 3 1 3 1
3 = 11i + 4j – k −2
4. Given p = 3i + 2k, q = i – 2j + 3k and r = –4i + 3j – k determine: (a) p × (r × q)
(b) (3p × 2r) × q
i j k (a) r × q = − 4 3 −1 = 7i + 11j + 5k 1 −2 3
Hence,
i j k p × (r × q) = 3 0 2 = (–22)i – (1)j + (33)k = –22i – j + 33k 7 11 5
(b) 3p × 2r =
i 9
j 0
k 6
= (–36)i – (–18 + 48)j + (54)k = –36i – 30j + 54k
− 8 6 −2 i j k (3p × 2r) × q = − 36 −30 54 = (–90 + 108)i – (–108 – 54)j + (72 + 30)k 1
−2
3
= 18i + 162j + 102k
5. For vectors p = 4i – j + 2k and q = –2i + 3j – 2k determine (i) p • q
(ii) p × q
(iii) p × q
(iv) q × p and (v) the angle between the vectors
(i)
p • q = (4)(– 2) + (–1)(3) +(2)(–2) = –15
(ii)
i j k p × q = 4 −1 2 = i ( 2 − 6 ) – j(– 8 + 4) + k(12 – 2) = −4i + 4 j + 10k − 2 3 −2
(iii) p ⋅ p = (4)(4) + (–1)(–1) + ( 2 )( 2 ) = 21 890
© 2014, John Bird
q ⋅ q = (–2)(–2) + (3)(3) + (–2)(–2) = 17 p ⋅ q = (4)(–2) + (–1)(3) + ( 2 )( −2 ) = –15 p×q =
= ( 357 − 225 )
2 ( 21)(17 ) − ( −15= )
i
132 = 11.49
k (iv) q × p = − 2 3 −2 = i ( 6 − 2 ) – j(– 4 + 8) + k(2 – 12) = 4i − 4 j − 10k 4 −1 2
(v)
cos θ =
j
(4)(−2) + (−1)(3) + ( 2 ) (−2) (4) 2 + (−1) 2 + ( 2 )
6. For vectors a = –7i + 4j +
(i)
(−2) 2 + (3) 2 + (−2) 2
=
−15 = −0.793884... 21 17
θ = cos −1 (−0.793884...) = 142.55°
from which,
(i) a • b
2
(ii) a × b
1 k and b = 6i – 5j – k find 2
(iii) a × b
(iv) b × a and (v) the angle between the vectors
1 1 a ⋅ b = (–7)(6) + (4)(–5) + ( )(–1) = –62 2 2
i (ii) a × b = − 7
6
j
k
5 1 1 = i −4 + – j(7 – 3) + k(35 – 24) = −1 i − 4 j + 11k 2 2 2 −5 −1 4
1 1 (iii) a ⋅ a = (–7)(–7) + (4)(4) + = 65.25 2 2 b ⋅ b = (6)(6) + (–5)(–5) + (–1)(–1) = 62
1 a ⋅ b = (–7)(6) + (4)(–5) + ( −1) = –62.5 2 a×b =
2 ( 65.25 )( 62 ) − ( −62.5 = )
i
(iv) b × a =
j
= ( 4045.5 − 3906.25 )
139.25 = 11.80
k
1 5 −5 −1 = i − + 4 – j(3 – 7) + k(24 – 35) = 1 i + 4 j − 11k 2 2 1 −7 4 2
6
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cos θ =
(v)
1 (−7)(6) + (4)(−5) + (−1) 2 1 (−7) 2 + (4) 2 + 2
2
=
(6) 2 + (−5) 2 + (−1) 2
−62.5 = −0.9826388 65.25 62
θ = cos −1 (−0.9826388) = 169.31°
from which,
7. Forces of (i + 3j), (– 2i – j), (i – 2j) newtons act at three points having position vectors of (2i + 5j), 4j and (–i + j) metres, respectively. Calculate the magnitude of the moment.
Magnitude of the moments = the sum of the magnitudes of the three forces When position vector, r = 2i + 5j and F = i + 3j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (2)(2) + (5)(5) +(0)(0) = 29 F ⋅ F = (1)(1) + (3)(3) +(0)(0) = 10 r ⋅ F = (2)(1) + (5)(3) + (0)(0) = 17
i.e.
M =
( 29 )(10 ) − (17 )
2
=1Nm
When position vector, r = 4j and F = –2i – j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (0)(0) + (4)(4) +(0)(0) = 16 F ⋅ F = (– 2)(–2) + (–1)(–1) +(0)(0) = 5 r ⋅ F = (0)(–2) + (4)(–1) + (0)(0) = –4
i.e.
M =
(16 )( 5) − ( −4 )
2
=8Nm
When position vector, r = –i + j and F = –i – 2j Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (1)(1) + (1)(1) + (0)(0) = 2 F ⋅ F = (–1)(–1) + (–2)(–2) + (0)(0) = 5 892
© 2014, John Bird
r ⋅ F = (1)(–1) + (1)(–2) + (0)(0) = –3 i.e.
M =
( 2 )( 5) − ( −3)
2
=1Nm
Hence, magnitude of the moment = 1 + 8 + 1 = 10 N m
8. A force of (2i – j + k) newtons acts on a line through point P having coordinates (0, 3, 1) metres. Determine the moment vector and its magnitude about point Q having coordinates (4, 0, –1) metres.
Position vector, r = (0i + 3j + k) – (4i + 0j – k) = –4i + 3j + 2k
Moment, M = r × F
i j k where M = − 4 3 2 = (3 + 2)i – (–4 – 4)j + (4 – 6)k 2 −1 1 = (5i + 8j – 2k) N m
Magnitude of M, M = r × F = where
( r ⋅ r )( F ⋅ F ) − ( r ⋅ F )
2
r ⋅ r = (–4)(–4) + (3)(3) + (2)(2) = 29 F ⋅ F = (2)(2) + (–1)(–1) +(1)(1) = 6 r ⋅ F = (–4)(2) + (3)(–1) +(2)(1) = –9
i.e.
M =
( 29 )( 6 ) − ( −9 )
2
= 9.64 N m
9. A sphere is rotating with angular velocity ω about the z-axis of a system, the axis coinciding with the axis of the sphere. Determine the velocity vector and its magnitude at position (–5i + 2j – 7k) m, when the angular velocity is (i + 2j) rad/s.
Velocity vector, v = ω × r = (i + 2j ) × (–5i + 2j – 7k) =
i
j
k
1
2
0
− 5 2 −7
= (–14 – 0)i – (– 7 – 0)j + (2 + 10)k = –14i + 7j + 12k
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Magnitude of v, v =
(ω ⋅ ω )( r ⋅ r ) − ( r ⋅ ω )
2
where ω ⋅ ω = (1)(1) + (2)(2) + (0)(0) = 5 r ⋅ r = (– 5)(–5) + (2)(2) + (–7)(–7) = 78 r ⋅ ω = (–5)(1) + (2)(2) + (–7)(0) = –1 Hence,
v =
( 5)( 78) − ( −1)
2
= 19.72 m/s
10. Calculate the velocity vector and its magnitude for a particle rotating about the z-axis at an angular velocity of (3i – j + 2k) rad/s when the position vector of the particle is at (i – 5j + 4k) m.
i j k Velocity vector, v = ω × r = (3i – j + 2k) × (i – 5j + 4k) = 3 1 2 1 −5 4
= (–4 + 10)i – (12 – 2)j + (–15 + 1)k = 6i – 10j – 14k Magnitude of v, v =
(ω ⋅ ω )( r ⋅ r ) − ( r ⋅ ω )
2
where ω ⋅ ω = (3)(3) + (–1)(–1) + (2)(2) = 14 r ⋅ r = (1)(1) + (–5)(–5) + (4)(4) = 42 r ⋅ ω = (1)(3) + (–5)(–1) + (4)(2) = 16 Hence,
v =
(14 )( 42 ) − (16 )
2
= 18.22 m/s
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EXERCISE 218 Page 596
1. Find the vector equation of the line through the point with position vector 5i – 2j + 3k which is parallel to the vector 2i + 7j – 4k. Determine the point on the line corresponding to λ = 2 in the resulting equation. Vector equation of the line, r = a + λb = (5i – 2j + 3k) + λ(2i + 7j – 4k) i.e.
r = (5 + 2λ)i + (7λ – 2)j + (3 – 4λ)k
When λ = 2,
r = 9i + 12j – 5k
2. Express the vector equation of the line in Problem 1 in standard Cartesian form.
The vector equation of a straight line in standard Cartesian form is:
x − a1 y − a2 z − a3 = = = λ b1 b2 b3 Since a = 5i – 2j + 3k,
a1 = 5, a2 = −2 and a3 = 3
and
b1 = 2, b2 = 7 and b3 = −4
b = 2i + 7j – 4k,
then the Cartesian equations are:
3. Express in vector form:
3x − 1 5 y + 1 4 − z = = 4 2 3
i.e.
i.e.
x −5 y + 2 z −3 = = = λ 2 7 −4
or
x −5 y + 2 3− z = = = λ 2 7 4
3x − 1 5 y + 1 4 − z = = 3 4 2
i.e.
1 1 3 x − 5 y + 3 5 −1( z − 4 ) = = 4 2 3
1 1 x− y+ 3 5 = = 4 2 3 5
( z − 4) −3
1 1 a1 = , a2 = − and a3 = 4 3 5 895
© 2014, John Bird
and
b1 =
4 2 , b2 = and b3 = −3 3 5
Hence, in vector form the equation is: r = ( a1 + λb1 ) i + ( a2 + λb2 ) j + ( a3 + λb3 ) k
1 4 = + λi + 3 3 or
r=
1 2 − + λ j + ( 4 − 3λ ) k 5 5
1 1 (1 + 4λ ) i + ( 2λ − 1) j + ( 4 − 3λ ) k 3 5
4. Express in vector form: 2x + 1 =
2 x + 1 1 − 4 y 3z − 1 = = 1 5 4
i.e.
3z − 1 1− 4 y = 4 5
1 1 1 z− x+ y− 3 2 = 4 = 1 5 4 − 2 4 3
i.e.
1 1 1 and a3 = a1 = − , a2 = 2 4 3
and
1 5 4 b1 = , b2 = − and b3 = 2 4 3
Hence, in vector form the equation is: r = ( a1 + λb1 ) i + ( a2 + λb2 ) j + ( a3 + λb3 ) k
1 5 1 1 1 4 = − + λi + − λj + + λk 4 4 2 2 3 3 or
r=
1 1 1 ( λ − 1) i + (1 − 5λ ) j + (1 + 4λ ) k 2 4 3
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