CS计算机代考程序代写 Multiple View Geometry: Solution Sheet 7

Multiple View Geometry: Solution Sheet 7
Prof. Dr. Florian Bernard, Florian Hofherr, Tarun Yenamandra
Computer Vision Group, TU Munich
Link Zoom Room , Password: 307238

Exercise: June 9th, 2021

Part I: Theory

1. (a) l is in the coimage of L, and therefore l is a normal vector to the plane that is determined
by the camera position and L.


lTx1 = 0
lTx2 = 0.

⇒ l ∼ x1 × x2 = x̂1×2.

l1 and l2 are normal vectors to the planes through camera position and L1, L2 respectively.


lT1 x = 0
lT2 x = 0

⇒ x ∼ l1 × l2 = l̂1l2.

(b) i. l1 ∼ x̂u :
x is in the preimage of L1. ⇒ l>1 x = 0.
∃ point u 6= p in L1. ⇒ l>1 u = 0
⇒ l1 ∼ x̂u.

ii. l2 ∼ x̂v : analog to i.
iii. x1 ∼ l̂r :

x1 is in the preimage of L. ⇒ x>1 l = 0
∃ a line L′ through p1 with coimage r 6= l. ⇒ x>1 r = 0.
⇒ x1 ∼ l̂r.

iv. x2 ∼ l̂s : analog to iii.

2. rank
(
x̂1Π1
x̂2Π2

)
5 3

⇒ ∃X ∈ R4\{0} with
(
x̂1Π1
x̂2Π2

)
X = 0.

⇒ x̂1Π1X = 0 ∧ x̂2Π2X = 0,

⇒ x1 ×Π1X = 0 ∧ x2 ×Π2X = 0.

⇒ x1 and Π1X are linearly dependent; and x2 and Π2X are linearly dependent.

⇒ ∃λ1, λ2 ∈ R with Π1X = λ1×1 ∧ Π2X = λ2×2

⇒ x1 and x2 are projections of X.

1

https://tum-conf.zoom.us/s/62772800235?pwd=SUpZN2QrV0JpeXJyR2R1TWx5cHEwdz09

3. ∃λ ∈ R : [R′, T ′] = λ [R, T ]H = λ [R, T ]
[
I 0
v> v4

]
= λ [R+ Tv>, T v4]

E′ = T̂ ′R′

= (λ̂v4T ) · (λ(R+ Tv>))
= λ2v4T̂ (R+ Tv

>)

= λ2v4T̂R+ λ
2v4 T̂ T︸︷︷︸

=0

v>

= λ2v4T̂R

= λ2v4E with λ
2v4 ∈ R

⇒ E′ ∼ E

2