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平面向量
共 3 道题目
2026-07-30
向量取值范围
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x
O
y
xOy
x
O
y
中
∣
O
A
→
∣
=
∣
O
B
→
∣
=
2
|\overrightarrow{OA}|=|\overrightarrow{OB}|=\sqrt{2}
∣
O
A
∣
=
∣
O
B
∣
=
2
,
∣
A
B
→
∣
=
2
|\overrightarrow{AB}|=2
∣
A
B
∣
=
2
,
C
(
3
,
4
)
C(3,4)
C
(
3
,
4
)
.
∣
2
C
A
→
+
A
B
→
∣
|2\overrightarrow{CA}+\overrightarrow{AB}|
∣2
C
A
+
A
B
∣
取值范围 ( )
A.
[
6
,
14
]
[6,14]
[
6
,
14
]
B.
[
6
,
12
]
[6,12]
[
6
,
12
]
C.
[
8
,
14
]
[8,14]
[
8
,
14
]
D.
[
8
,
12
]
[8,12]
[
8
,
12
]
2026-07-24
向量与分段函数
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f
(
x
)
=
{
1
,
x
>
0
0
,
x
=
0
−
1
,
x
<
0
f(x)=\begin{cases}1,x>0\\0,x=0\\-1,x<0\end{cases}
f
(
x
)
=
⎩
⎨
⎧
1
,
x
>
0
0
,
x
=
0
−
1
,
x
<
0
,
a
⃗
,
b
⃗
,
c
⃗
\vec{a},\vec{b},\vec{c}
a
,
b
,
c
为平面内三个不同单位向量.
f
(
a
⃗
⋅
b
⃗
)
+
f
(
b
⃗
⋅
c
⃗
)
+
f
(
c
⃗
⋅
a
⃗
)
=
0
f(\vec{a}\cdot\vec{b})+f(\vec{b}\cdot\vec{c})+f(\vec{c}\cdot\vec{a})=0
f
(
a
⋅
b
)
+
f
(
b
⋅
c
)
+
f
(
c
⋅
a
)
=
0
,
∣
a
⃗
+
b
⃗
+
c
⃗
∣
|\vec{a}+\vec{b}+\vec{c}|
∣
a
+
b
+
c
∣
取值范围
‾
\underline{\qquad}
.
2026-07-02
向量线性组合
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已知
∣
a
⃗
∣
=
a
⃗
⋅
b
⃗
=
1
|\vec{a}|=\vec{a}\cdot\vec{b}=1
∣
a
∣
=
a
⋅
b
=
1
,
∣
b
⃗
∣
>
1
|\vec{b}|>1
∣
b
∣
>
1
,记
c
⃗
=
λ
a
⃗
+
μ
b
⃗
\vec{c}=\lambda\vec{a}+\mu\vec{b}
c
=
λ
a
+
μ
b
。当
a
⃗
+
b
⃗
−
c
⃗
=
0
⃗
\vec{a}+\vec{b}-\vec{c}=\vec{0}
a
+
b
−
c
=
0
时,
λ
+
μ
=
\lambda+\mu=
λ
+
μ
=
‾
\underline{\qquad}
,当
∣
a
⃗
+
b
⃗
−
c
⃗
∣
=
1
|\vec{a}+\vec{b}-\vec{c}|=1
∣
a
+
b
−
c
∣
=
1
时,
λ
+
μ
\lambda+\mu
λ
+
μ
的取值范围为
‾
\underline{\qquad}
。