平面向量
(1)求|
(2)是否存在实数t,使
|
(1)|
|=a
=362+(-3)2 5
|
|=b
=1+22
,5
•a
=(6,-3)(1,2)=6-6=0b
(2)∵
⊥x y
∴
•x
=0y
即
•x
=[y
+(t-6)a
] (b
+ta
)=|b
|2+t(t-6)|a
|2=45+5t(t-6)=0b
解得t=3
∴存在t=3使得
⊥x
.y
平面向量
(1)求|
(2)是否存在实数t,使
|
(1)|
|=a
=362+(-3)2 5
|
|=b
=1+22
,5
•a
=(6,-3)(1,2)=6-6=0b
(2)∵
⊥x y
∴
•x
=0y
即
•x
=[y
+(t-6)a
] (b
+ta
)=|b
|2+t(t-6)|a
|2=45+5t(t-6)=0b
解得t=3
∴存在t=3使得
⊥x
.y