|
设已知两点
M
1
(
5
,
2
,
2
)
,
M
2
(
4
,
0
,
3
)
M_1(5, \sqrt{2}, 2), M_2(4, 0, 3)
M1(5,2
,2),M2(4,0,3) ，计算向量
M
1
M
2
→
\overrightarrow{M_1M_2}
M1M2
模长
M
1
M
2
→
=
M
2
→
−
M
1
→
=
(
−
1
,
−
2
,
1
)
∣
M
1
M
2
→
∣
=
(
−
1
)
2
+
(
−
2
)
2
+
1
2
=
2
\overrightarrow{M_1M_2}=\overrightarrow{M_2}-\overrightarrow{M_1}=(-1, -\sqrt{2}, 1) \\ |\overrightarrow{M_1M_2}|=\sqrt{(-1)^2 + (-\sqrt{2})^2 + 1^2} = 2
M1M2
=M2
−M1
=(−1,−2
,1)∣M1M2
∣=(−1)2+(−2
)2+12
=2方向余弦
c
o
s
α
=
x
∣
M
1
M
2
→
∣
=
−
1
2
=
−
1
2
c
o
s
β
=
y
∣
M
1
M
2
→
∣
=
−
2
2
=
−
2
2
c
o
s
γ
=
x
∣
M
1
M
2
→
∣
=
1
2
=
1
2
cos\alpha = \frac{x}{|\overrightarrow{M_1M_2}|} = \frac{-1}{2} = - \frac{1}{2} \\ cos\beta = \frac{y}{|\overrightarrow{M_1M_2}|} = \frac{-\sqrt{2}}{2} = - \frac{\sqrt{2}}{2} \\ cos\gamma = \frac{x}{|\overrightarrow{M_1M_2}|} = \frac{1}{2} = \frac{1}{2} \\
cosα=∣M1M2
∣x=2−1=−21cosβ=∣M1M2
∣y=2−2
=−22
cosγ=∣M1M2
∣x=21=21 其中：
(
c
o
s
α
)
2
+
(
c
o
s
β
)
2
+
(
c
o
s
γ
)
2
=
1
(cos\alpha)^2 + (cos\beta)^2 + (cos\gamma)^2 = 1
(cosα)2+(cosβ)2+(cosγ)2=1方向角
α
=
2
π
3
,
β
=
3
π
4
,
γ
=
π
3
\alpha = \frac{2\pi}{3}, \beta = \frac{3\pi}{4}, \gamma = \frac{\pi}{3}\\
α=32π,β=43π,γ=3π方向一致的单位向量
M
1
M
2
→
∣
M
1
M
2
→
∣
=
(
−
1
,
−
2
,
1
)
2
=
(
−
1
2
,
−
2
2
,
1
2
)
\frac{\overrightarrow{M_1M_2}}{|\overrightarrow{M_1M_2}|} = \frac{(-1, -\sqrt{2}, 1)}{2} = (-\frac{1}{2}, -\frac{\sqrt{2}}{2}, \frac{1}{2}) \\
∣M1M2
∣M1M2
=2(−1,−2
,1)=(−21,−22
,21)设
f
(
x
,
y
,
z
)
=
x
+
y
2
+
z
3
f(x, y, z) = x + y^2 + z^3
f(x,y,z)=x+y2+z3，求
f
f
f 在点
P
0
(
1
,
1
,
1
)
P_0(1, 1, 1)
P0(1,1,1) 沿方向
M
1
M
2
→
\overrightarrow{M_1M_2}
M1M2
的方向导数。 解：易见
f
f
f 在
P
0
(
1
,
1
,
1
)
P_0(1, 1, 1)
P0(1,1,1) 可微，所以：
f
x
(
P
0
)
=
1
,
f
y
(
P
0
)
=
2
,
f
z
(
P
0
)
=
3
f_x(P_0) = 1, f_y(P_0) = 2, f_z(P_0) = 3 \\
fx(P0)=1,fy(P0)=2,fz(P0)=3 根据 2. 方向余弦，因此方向导数
f
M
1
M
2
→
f_{\overrightarrow{M_1M_2}}
fM1M2
：
f
M
1
M
2
→
(
P
0
)
=
f
x
(
P
0
)
⋅
c
o
s
α
+
f
y
(
P
0
)
⋅
c
o
s
β
+
f
z
(
P
0
)
⋅
c
o
s
γ
=
1
∗
(
−
1
2
)
+
2
∗
(
−
2
2
)
+
3
∗
1
2
f_{\overrightarrow{M_1M_2}}(P_0) = f_x(P_0) · cos\alpha + f_y(P_0) · cos\beta+ f_z(P_0) · cos\gamma \\ = 1 *( - \frac{1}{2}) + 2 * (- \frac{\sqrt{2}}{2}) + 3 * \frac{1}{2}
fM1M2
(P0)=fx(P0)⋅cosα+fy(P0)⋅cosβ+fz(P0)⋅cosγ=1∗(−21)+2∗(−22
)+3∗21
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