精编《力学》杜婵英 漆安慎课后习题答案,帮你轻松掌握力学
解：⑴(AK+BK CK)×CK+(CK+AK+BK)×AK+(AK BK+CK)×BK
=AK×CK+BK×CK+CK×AK+BK×AK+AK×BK+CK×BK
=0
⑵i ×( j+k ) j×(i +k )+k ×(i + j+k ) j
=k j+k i + j i
=2k 2i i k
⑶(2AK+BK)×(CK AK)+(BK+CK
)×(AK+BK)
=2AK×(CK AK)+BK×KK
KKKKKK=2AK×CK+BK×CK BK(C×AK A)+BK+B×(A+B)+C×(A+B)
×AK+CK×AK+CK×B
K
=AK×C
K
14．计算下面诸式
解：⑴i ( j×k )+k (i × j)+ j (k ×i ) j
i =i i +k k + j j=3
k
⑵AK (BK×AK)=BK (AK×AK
)=0
15．求证：(AK+BK) [(AK+CK)×BK)]= AK (BK×CK
)
证明：(AK+BK) [(AK+CK)×BK
)]
=AK (AK×BK+CK×BK)+BK (KKKK=AK (AK×BK)+AKKKAK×BK+CK×B)KKK=BK (AK×AK
)+AK (C (CK×B)+B (A×B)+B (C×B)
×BK)+AK (BK×BK)+CK (BKK =AK (CK×BK)= AK (BK×CK×B))
K16．已知AK=(1+2t2)i +e t j k ，求dAd2AK,dt2.
解：
dAK=ddt
dt[(1+2t2)i +e t j k ]=4ti e t j d2AK
dt=
d(4ti t e j)=4i +e t j
17．已知AK
=3e ti (4t3 t) j+tk ,BK
=4t2i +3t j， 求dt
(AK BK
)
解：AK BK
=AxBx+AyBy+AzBz t
=3e 4t2 3t(4t3 t)=12t2e t 12t4+3t
2
(AK BK)=
dt
dt
(12t2e t 12t4+3t2)
=12(2t t2
)e t
48t3+6t