Pou menm non ak atik la, gade Distans .
Distans ki soti nan pwen A ak plan P se AH. Distans sa a mwens pase AM ak AM' Nan espas eklidyen , distans ant yon pwen ak yon plan se distans ki pi kout ki separe pwen sa a ak yon pwen sou plan an. Teyorèm Pitagò a pèmèt nou deklare distans ki genyen ant pwen A ak plan an (P) koresponn ak distans ki separe A ak pwojeksyon òtogonal li H sou plan an (P).
Si yo bay espas la yon ankadreman referans ortonormal , yo ka defini pwen yo lè l sèvi avèk kowòdone yo yo rele kowòdone katezyen.
Swa nan espas:
Pwen A ak kowòdone
(
x
a
,
y
a
,
z
a
)
{\displaystyle (x_{a},y_{a},z_{a})}
Nenpòt pwen M nan plan P la
Pwojeksyon ortogonal H nan A sou P, te note
H
(
x
h
,
y
h
,
z
h
)
{\displaystyle H(x_{h},y_{h},z_{h})}
Plan P ekwasyon katezyen an: ax + by + cz + d = 0
n
→
(
a
b
c
)
{\displaystyle {\vec {n}}{\begin{pmatrix}a\\b\\c\\\end{pmatrix}}}
vektè nòmal nan plan P laLè sa a, distans
δ
{\displaystyle \delta }
δ
A
,
P
{\displaystyle \delta _{\mathrm {A} ,\mathrm {P} }}
δ
A
,
P
=
|
n
→
⋅
M
A
→
|
|
|
n
→
|
|
{\displaystyle \delta _{\mathrm {A} ,\mathrm {P} }={\frac {\left|{\vec {n}}\cdot {\vec {MA}}\right|}{||{\vec {n}}||}}}
pakonsekan,
δ
A
,
P
=
|
a
x
a
+
b
y
a
+
c
z
a
+
d
|
a
2
+
b
2
+
c
2
{\displaystyle \delta _{\mathrm {A} ,\mathrm {P} }={\frac {\left|ax_{a}+by_{a}+cz_{a}+d\right|}{\sqrt {a^{2}+b^{2}+c^{2}}}}}
Demonstrasyon Premyèman, nou konnen vektè
A
H
→
{\displaystyle {\overrightarrow {\mathrm {AH} }}}
n
→
{\displaystyle {\vec {n}}}
A
H
→
=
λ
⋅
n
→
{\displaystyle {\overrightarrow {\mathrm {AH} }}=\lambda \cdot {\vec {n}}}
ki se,
(
x
h
−
x
a
y
h
−
y
a
z
h
−
z
a
)
=
λ
(
a
b
c
)
{\displaystyle {\begin{pmatrix}x_{h}-x_{a}\\y_{h}-y_{a}\\z_{h}-z_{a}\\\end{pmatrix}}=\lambda {\begin{pmatrix}a\\b\\c\\\end{pmatrix}}}
H
∈
P
{\displaystyle \mathrm {H} \in \mathrm {P} }
a
x
h
+
b
y
h
+
c
z
h
+
d
=
0
{\displaystyle ax_{h}+by_{h}+cz_{h}+d=0}
{
x
h
=
λ
a
+
x
a
y
h
=
λ
b
+
y
a
z
h
=
λ
c
+
z
a
a
x
h
+
b
y
h
+
c
z
h
+
d
=
0
{\displaystyle \left\{{\begin{matrix}x_{h}=\lambda a+x_{a}\\y_{h}=\lambda b+y_{a}\\z_{h}=\lambda c+z_{a}\\ax_{h}+by_{h}+cz_{h}+d=0\end{matrix}}\right.}
Ranplase kowòdone H yo nan 4yèm ekwasyon pa valè yo jwenn nan premye 3 yo pèmèt nou ekri:
a
(
λ
a
+
x
a
)
+
b
(
λ
b
+
y
a
)
+
c
(
λ
c
+
z
a
)
+
d
=
0
{\displaystyle a(\lambda a+x_{\mathrm {a} })+b(\lambda b+y_{\mathrm {a} })+c(\lambda c+z_{\mathrm {a} })+d=0}
oswa:
a
x
a
+
b
y
a
+
c
z
a
+
d
+
λ
(
a
2
+
b
2
+
c
2
)
=
0
{\displaystyle ax_{a}+by_{a}+cz_{a}+d+\lambda (a^{2}+b^{2}+c^{2})=0}
P se yon plan, a, b, c pa tout zewo: nou genyen
λ
=
−
a
x
a
+
b
y
a
+
c
z
a
+
d
a
2
+
b
2
+
c
2
{\displaystyle \lambda =-{\frac {ax_{a}+by_{a}+cz_{a}+d}{a^{2}+b^{2}+c^{2}}}}
Finalman, distans ki soti nan A rive nan P se pa lòt ke longè vektè
A
H
→
{\displaystyle {\overrightarrow {\mathrm {AH} }}}
δ
A
,
P
=
‖
A
H
→
‖
=
|
λ
|
‖
n
→
‖
{\displaystyle \delta _{\mathrm {A} ,\mathrm {P} }=\lVert {\vec {AH}}\rVert =\left|\lambda \right|\lVert {\vec {n}}\rVert }
kite
δ
A
,
P
=
|
−
(
a
x
a
+
b
y
a
+
c
z
a
+
d
)
a
2
+
b
2
+
c
2
|
a
2
+
b
2
+
c
2
{\displaystyle \delta _{\mathrm {A} ,\mathrm {P} }=\left|{\frac {-(ax_{a}+by_{a}+cz_{a}+d)}{a^{2}+b^{2}+c^{2}}}\right|{\sqrt {a^{2}+b^{2}+c^{2}}}}
epi finalman
δ
A
,
P
=
|
a
x
a
+
b
y
a
+
c
z
a
+
d
|
a
2
+
b
2
+
c
2
{\displaystyle \delta _{\mathrm {A} ,P}={\frac {\left|ax_{a}+by_{a}+cz_{a}+d\right|}{\sqrt {a^{2}+b^{2}+c^{2}}}}}
Sa a konplete prèv la.
