问题
填空题
若|
|
答案
∵
⊥a b
∴
•a
=0b
∵(2
+3a
)⊥(kb
-4a
)b
∴∵(2
+3a
)•(kb
-4a
)=0b
即2k
2+(3k-8)a
•a
-12b
2=0b
∵|
|=|a
|=1b
∴2k-12=0
∴k=6
故答案为6
若|
|
∵
⊥a b
∴
•a
=0b
∵(2
+3a
)⊥(kb
-4a
)b
∴∵(2
+3a
)•(kb
-4a
)=0b
即2k
2+(3k-8)a
•a
-12b
2=0b
∵|
|=|a
|=1b
∴2k-12=0
∴k=6
故答案为6