한 면만 보이는 곳

[Graphics:../HTMLFiles/index_121.gif]

각 V1, V가 원점을 지나는 직선 표기

V1→ {V1x == Dv1 Cos[θ], V1y == Dv1 Sin[θ]}

V→ {Vx == -Dv Sin[θ], Vy == Dv Cos[θ]}

V, V1, (Bx,By) 을 동시에 지나는 직선

By == (V1y - Vy)/(V1x - Vx) (Bx - Vx) + Vy

DvH1 : Bz = V1~V 거리 : B~V 거리

DvH1/Bz == ((V1x - Vx)^2 + (V1y - Vy)^2)^(1/2)/((Bx - Vx)^2 + (By - Vy)^2)^(1/2)

Cos[θ],Sin[θ] 값 구하기

-(Dv^2 + Dv1^2) Bz^2 + DvH1^2 (Bx^2 + By^2 + Dv^2 - 2 By Dv Cos[θ] + 2 Bx Dv Sin[θ]) == 0

-Dv Dv1 + (Bx Dv + By Dv1) Cos[θ] + (By Dv - Bx Dv1) Sin[θ] == 0

Cos[θ] → (-Bz^2 (By Dv - Bx Dv1) (Dv^2 + Dv1^2) + (By Dv (Bx^2 + By^2 + Dv^2) - Bx (Bx^2 + By^2 - Dv^2) Dv1) DvH1^2)/(2 (Bx^2 + By^2) Dv^2 DvH1^2)

Sin[θ] → (Bz^2 (Bx Dv + By Dv1) (Dv^2 + Dv1^2) - (Bx Dv (Bx^2 + By^2 + Dv^2) + By (Bx^2 + By^2 - Dv^2) Dv1) DvH1^2)/(2 (Bx^2 + By^2) Dv^2 DvH1^2)

V : (Vx, Vy)

Vx→ (-Bz^2 (Bx Dv + By Dv1) (Dv^2 + Dv1^2) + (Bx Dv (Bx^2 + By^2 + Dv^2) + By (Bx^2 + By^2 - Dv^2) Dv1) DvH1^2)/(2 (Bx^2 + By^2) Dv DvH1^2)

Vy→ (Bz^2 (-By Dv + Bx Dv1) (Dv^2 + Dv1^2) + (By Dv (Bx^2 + By^2 + Dv^2) - Bx (Bx^2 + By^2 - Dv^2) Dv1) DvH1^2)/(2 (Bx^2 + By^2) Dv DvH1^2)

Solve

Source