问题
填空题
已知{
|
答案
∵{
,i
,j
}是单位正交基底,k
∴
=-3a
+4i
-j
=(-3,4,-1),k
-a
=-8b
+16i
-3j
=(-8,16,-3)k
由此可得
=b
-(a
-a
)=(-3,4,-1)-(-8,16,-3)=(5,-12,2).b
∴
•a
=-3×5+4×(-12)+(-1)×2=-65.b
故答案为:-65
已知{
|
∵{
,i
,j
}是单位正交基底,k
∴
=-3a
+4i
-j
=(-3,4,-1),k
-a
=-8b
+16i
-3j
=(-8,16,-3)k
由此可得
=b
-(a
-a
)=(-3,4,-1)-(-8,16,-3)=(5,-12,2).b
∴
•a
=-3×5+4×(-12)+(-1)×2=-65.b
故答案为:-65