题目内容

已知△ABC中,A(2,-5,3),
AB
=(4,1,2),
BC
=(3,-2,5).
(1)求其余顶点的坐标;
(2)求
CA

(3)求角A.
分析:(1)利用
OB
=
OA
+
AB
OC
=
OB
+
BC
,即可得出.
(2)利用
CA
=
CB
+
BA
即可得出.
(3)利用向量的夹角公式即可得出.
解答:解:(1)∵在△ABC中,A(2,-5,3),
AB
=(4,1,2),
BC
=(3,-2,5).
OB
=
OA
+
AB
=(2,-5,3)+(4,1,2)=(6,-4,5),
OC
=
OB
+
BC
=(6,-4,5)+(3,-2,5)=(9,-6,10).
(2)
CA
=
CB
+
BA
=(-3,2,-5)+(-4,-1,-2)=(-7,1,-7).
(3)∵
AB
AC
=(4,1,2)•(7,-1,7)=28-1+14=41.
|
AB
|=
42+12+22
=
21
|
AC
|=
72+(-1)2+72
=
99

∴cos∠A=
AB
AC
|
AB
| |
AC
|
=
41
21
99
=
41
231
693
点评:本题考查了向量的加法运算和向量的夹角公式,属于基础题.
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