📐 Distance d’un point à un plan
Soit ( P ) (P) ( P ) A A A H H H A A A ( P ) (P) ( P )
La distance A H AH A H A A A ( P ) (P) ( P ) d ( A , ( P ) ) = A H d(A,(P))=AH d ( A , ( P )) = A H
Prenons : A ( x A ; y A ; z A ) A(x_A;y_A;z_A) A ( x A ; y A ; z A ) n ⃗ ( a ; b ; c ) \vec{n}(a;b;c) n ( a ; b ; c )
Le plan ( P ) (P) ( P )
a x + b y + c z + d = 0 ax+by+cz+d=0 a x + b y + cz + d = 0 on a :
H A → ( x A − x H y A − y H z A − z H ) \overrightarrow{HA}\begin{pmatrix}x_A-x_H \\ y_A-y_H \\ z_A-z_H\end{pmatrix} H A ⎝ ⎛ x A − x H y A − y H z A − z H ⎠ ⎞ Donc
H A → ⋅ n ⃗ = a ( x A − x H ) + b ( y A − y H ) + c ( z A − z H ) = a x A + b y A + c z A − ( a x H + b y H + c z H ) \begin{align*}
\overrightarrow{HA}\cdot\vec{n}&=a(x_A-x_H)+b(y_A-y_H)+c(z_A-z_H)\\
&=ax_A+by_A+cz_A-(ax_H+by_H+cz_H)
\end{align*} H A ⋅ n = a ( x A − x H ) + b ( y A − y H ) + c ( z A − z H ) = a x A + b y A + c z A − ( a x H + b y H + c z H ) Puisque : H ∈ ( P ) H\in\left(P\right) H ∈ ( P ) a x H + b y H + c z H = − d ax_H+by_H+cz_H=-d a x H + b y H + c z H = − d
Ainsi : ∣ H A → ⋅ n ⃗ ∣ = ∣ a x A + b y A + c z A + d ∣ ( 1 ) \left|\overrightarrow{HA}\cdot\vec{n}\right|=\left|ax_A+by_A+cz_A+d\right| ~~~~ (1) ∣ ∣ H A ⋅ n ∣ ∣ = ∣ a x A + b y A + c z A + d ∣ ( 1 )
et on a : ∣ H A → ⋅ n ⃗ ∣ = H A . ∥ n ⃗ ∥ . ∣ c o s ( H A → ; n ⃗ ‾ ) ∣ = H A . ∥ n ⃗ ∥ |\overrightarrow{HA}\cdot\vec{n}|= HA.\|\vec{n}\|.\left|cos\left(\overline{\overrightarrow{HA};\vec{n}}\right)\right|=HA.\|\vec{n}\| ∣ H A ⋅ n ∣ = H A .∥ n ∥. ∣ ∣ cos ( H A ; n ) ∣ ∣ = H A .∥ n ∥
car : ∣ c o s ( H A → ; n ⃗ ‾ ) ∣ = ∣ ± 1 ∣ = 1 \left|cos\left(\overline{\overrightarrow{HA};\vec{n}}\right)\right|=|\pm1|=1 ∣ ∣ cos ( H A ; n ) ∣ ∣ = ∣ ± 1∣ = 1
, Alors : ∣ H A → ⋅ n ⃗ ∣ = H A . ∥ n ⃗ ∥ ( 2 ) |\overrightarrow{HA}\cdot\vec{n}|=HA.\|\vec{n}\| ~~~~ (2) ∣ H A ⋅ n ∣ = H A .∥ n ∥ ( 2 )
De ( 1 ) (1) ( 1 ) ( 2 ) (2) ( 2 )
A H = ∣ a x A + b y A + c z A + d ∣ ∥ n ⃗ ∥ = ∣ a x A + b y A + c z A + d ∣ a 2 + b 2 + c 2 AH=\dfrac{\left|ax_A+by_A+cz_A+d\right|}{\|\vec{n}\|}
=\dfrac{\left|ax_A+by_A+cz_A+d\right|}{\sqrt{a^2+b^2+c^2}} A H = ∥ n ∥ ∣ a x A + b y A + c z A + d ∣ = a 2 + b 2 + c 2 ∣ a x A + b y A + c z A + d ∣
Propriété
Si : ( P ) : a x + b y + c z + d = 0 \left(P\right):ax+by+cz+d=0 ( P ) : a x + b y + cz + d = 0 A ( x A ; y A ; z A ) A\left(x_A;y_A;z_A\right) A ( x A ; y A ; z A )
d ( A ; ( P ) ) = ∣ a x A + b y A + c z A + d ∣ a 2 + b 2 + c 2 d\left(A;\left(P\right)\right)=\dfrac{\left|ax_A+by_A+cz_A+d\right|}{\sqrt{a^2+b^2+c^2}} d ( A ; ( P ) ) = a 2 + b 2 + c 2 ∣ a x A + b y A + c z A + d ∣
🚦Intersection plan sphère
Soient ( P ) (P) ( P ) ( S ) (S) ( S ) Ω \Omega Ω R R R d = d ( Ω , ( P ) ) d=d(\Omega,(P)) d = d ( Ω , ( P ))
Si d ( Ω , ( P ) ) = R d(\Omega,(P))=R d ( Ω , ( P )) = R ( P ) (P) ( P ) ( S ) (S) ( S ) H H H ( P ) (P) ( P ) ( S ) (S) ( S )
Si d ( Ω , ( P ) ) > R d(\Omega,(P))>R d ( Ω , ( P )) > R ( P ) (P) ( P ) ( S ) (S) ( S )
Si d ( Ω , ( P ) ) < R d(\Omega,(P))<R d ( Ω , ( P )) < R ( P ) (P) ( P ) ( S ) (S) ( S ) ( C ) (C) ( C ) r = R 2 − d 2 r=\sqrt{R^2-d^2} r = R 2 − d 2 H H H ( P ) (P) ( P ) Ω \Omega Ω ( P ) (P) ( P )
🔢 Application
Étudier : l’intersection de ( P ) (P) ( P ) S ( Ω , R ) S(\Omega,R) S ( Ω , R )
( P ) : 2 x − 2 y + z + 3 = 0 ; ; S ( Ω ; R ) : x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5 \left(P\right):2x-2y+z+3=0 ~~~;;~~~ S\left(\Omega;R\right):x^2+y^2+z^2-2x+6y+2z=5 ( P ) : 2 x − 2 y + z + 3 = 0 ;; S ( Ω ; R ) : x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5
Cherchons Ω \Omega Ω R R R M ( x ; y ; z ) ∈ S ( Ω , R ) M(x;y;z)\in S(\Omega,R) M ( x ; y ; z ) ∈ S ( Ω , R )
x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5 x^2+y^2+z^2-2x+6y+2z=5 x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5 On a : a = − 2 a=-2 a = − 2 b = 6 b=6 b = 6 c = 2 c=2 c = 2 d = − 5 d=-5 d = − 5 Δ = a 2 + b 2 + c 2 − 4 d = 64 \Delta = a^2 + b^2 + c^2 - 4d = 64 Δ = a 2 + b 2 + c 2 − 4 d = 64 ( S ) (S) ( S ) Ω ( − a 2 , − b 2 , − c 2 ) \Omega\left(-\dfrac{a}{2}, -\dfrac{b}{2}, -\dfrac{c}{2}\right) Ω ( − 2 a , − 2 b , − 2 c ) Ω ( 1 ; − 3 ; − 1 ) \Omega(1; -3; -1) Ω ( 1 ; − 3 ; − 1 ) R = Δ 2 = 4 R = \dfrac{\sqrt{\Delta}}{2} = 4 R = 2 Δ = 4
Autre méthode :
M ( x ; y ; z ) ∈ S ( Ω , R ) ⟺ x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5 ⟺ x 2 − 2 x + 1 + y 2 + 6 y + 9 + z 2 + 2 z + 1 = 5 + 1 + 9 + 1 ⟺ ( x − 1 ) 2 + ( y + 3 ) 2 + ( z + 1 ) 2 = 4 2 \begin{aligned}
M(x;y;z)\in S(\Omega,R) &\Longleftrightarrow x^2 + y^2 + z^2 - 2x + 6y + 2z = 5 \\
&\Longleftrightarrow x^2 - 2x + 1 + y^2 + 6y + 9 + z^2 + 2z + 1 = 5 + 1 + 9 + 1 \\
&\Longleftrightarrow (x - 1)^2 + (y + 3)^2 + (z + 1)^2 = 4^2
\end{aligned} M ( x ; y ; z ) ∈ S ( Ω , R ) ⟺ x 2 + y 2 + z 2 − 2 x + 6 y + 2 z = 5 ⟺ x 2 − 2 x + 1 + y 2 + 6 y + 9 + z 2 + 2 z + 1 = 5 + 1 + 9 + 1 ⟺ ( x − 1 ) 2 + ( y + 3 ) 2 + ( z + 1 ) 2 = 4 2 Donc Ω ( 1 ; − 3 ; − 1 ) \Omega(1;-3;-1) Ω ( 1 ; − 3 ; − 1 ) R = 4 R=4 R = 4
Calculons d ( Ω , ( P ) ) d(\Omega,(P)) d ( Ω , ( P ))
d ( Ω , ( P ) ) = ∣ a x Ω + b y Ω + c z Ω + d ∣ a 2 + b 2 + c 2 = ∣ 2 ⋅ 1 − 2 ⋅ ( − 3 ) + 1 ⋅ ( − 1 ) + 3 ∣ 2 2 + ( − 2 ) 2 + 1 2 = 10 3 \begin{aligned}
d(\Omega,(P)) &= \dfrac{|a x_\Omega + b y_\Omega + c z_\Omega + d|}{\sqrt{a^2 + b^2 + c^2}}\\
&= \dfrac{|2 \cdot 1 - 2 \cdot (-3) + 1 \cdot (-1) + 3|}{\sqrt{2^2 + (-2)^2 + 1^2}}\\
&= \dfrac{10}{3}
\end{aligned} d ( Ω , ( P )) = a 2 + b 2 + c 2 ∣ a x Ω + b y Ω + c z Ω + d ∣ = 2 2 + ( − 2 ) 2 + 1 2 ∣2 ⋅ 1 − 2 ⋅ ( − 3 ) + 1 ⋅ ( − 1 ) + 3∣ = 3 10 Or d ( Ω , ( P ) ) < R d(\Omega,(P)) < R d ( Ω , ( P )) < R ( P ) (P) ( P ) ( S ) (S) ( S ) ( C ) (C) ( C ) H H H
r = R 2 − d 2 = 4 2 − ( 10 3 ) 2 = 44 3 r = \sqrt{R^2 - d^2} = \sqrt{4^2 - \left(\dfrac{10}{3}\right)^2} = \dfrac{\sqrt{44}}{3} r = R 2 − d 2 = 4 2 − ( 3 10 ) 2 = 3 44
Cherchons les coordonnées du point H H H ( C ) (C) ( C ) ( Δ ) (\Delta) ( Δ ) H H H Ω \Omega Ω n ⃗ ( 2 ; − 2 ; 1 ) \vec{n}(2; -2; 1) n ( 2 ; − 2 ; 1 ) ( P ) (P) ( P ) ( Δ ) ⊥ ( P ) (\Delta) \perp (P) ( Δ ) ⊥ ( P ) n ⃗ \vec{n} n ( Δ ) (\Delta) ( Δ )
Une représentation paramétrique de ( Δ ) (\Delta) ( Δ )
{ x = 1 + 2 t y = − 3 − 2 t z = − 1 + t ( t ∈ R ) \left\{
\begin{array}{l}
x = 1 + 2t \\
y = -3 - 2t \\
z = -1 + t \quad (t \in \mathbb{R})
\end{array}
\right. ⎩ ⎨ ⎧ x = 1 + 2 t y = − 3 − 2 t z = − 1 + t ( t ∈ R ) H H H ( P ) (P) ( P ) ( Δ ) (\Delta) ( Δ ) H H H
{ x = 1 + 2 t y = − 3 − 2 t z = − 1 + t 2 x − 2 y + z + 3 = 0 \left\{
\begin{array}{l}
x = 1 + 2t \\
y = -3 - 2t \\
z = -1 + t \\
2x - 2y + z + 3 = 0
\end{array}
\right. ⎩ ⎨ ⎧ x = 1 + 2 t y = − 3 − 2 t z = − 1 + t 2 x − 2 y + z + 3 = 0 Remplaçons dans l’équation du plan :
2 ( 1 + 2 t ) − 2 ( − 3 − 2 t ) + ( − 1 + t ) + 3 = 0 ⇒ 2 + 4 t + 6 + 4 t − 1 + t + 3 = 0 ⇒ 9 t = − 10 ⇒ t = − 10 9 2(1 + 2t) - 2(-3 - 2t) + (-1 + t) + 3 = 0 \\
\Rightarrow 2 + 4t + 6 + 4t - 1 + t + 3 = 0 \\
\Rightarrow 9t = -10 \Rightarrow t = -\dfrac{10}{9} 2 ( 1 + 2 t ) − 2 ( − 3 − 2 t ) + ( − 1 + t ) + 3 = 0 ⇒ 2 + 4 t + 6 + 4 t − 1 + t + 3 = 0 ⇒ 9 t = − 10 ⇒ t = − 9 10 D’où :
{ x = 1 + 2 ⋅ ( − 10 9 ) = − 11 9 y = − 3 − 2 ⋅ ( − 10 9 ) = − 7 9 z = − 1 + ( − 10 9 ) = − 19 9 \left\{
\begin{array}{l}
x = 1 + 2 \cdot \left(-\dfrac{10}{9}\right) = -\dfrac{11}{9} \\
y = -3 - 2 \cdot \left(-\dfrac{10}{9}\right) = -\dfrac{7}{9} \\
z = -1 + \left(-\dfrac{10}{9}\right) = -\dfrac{19}{9}
\end{array}
\right. ⎩ ⎨ ⎧ x = 1 + 2 ⋅ ( − 9 10 ) = − 9 11 y = − 3 − 2 ⋅ ( − 9 10 ) = − 9 7 z = − 1 + ( − 9 10 ) = − 9 19 Alors H ( − 11 9 ; − 7 9 ; − 19 9 ) H\left(-\dfrac{11}{9}; -\dfrac{7}{9}; -\dfrac{19}{9}\right) H ( − 9 11 ; − 9 7 ; − 9 19 )