Computer Vision Week 2 NPTEL Assignment Answers 2025
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✅ Subject: Computer Vision
📅 Week: 2
🎯 Session: NPTEL 2025 July-October
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NPTEL Computer Vision Week 2 Assignment Answers 2025
1. Consider two points in projective space p 1 (9,27,3) and p2(28, 49, 7). Find the line passing through the points pl and p2 in 2D-space.
a) 2x – y – 15 = 0
b) 2x – y + 15 = 0
c) 2x + y – 15 = 0
d) 2x + y + 15 = 0
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2. Compute the transformed point of intersection of the straight lines-4x + 4y + 5 = 0 and x + 2y – 2 = 0 in R2.
a) p(-7, -8, 2)
b) p(28, -8,8)
c) p(-7, -32,2)
d) p(-7,8, 8)
Answer :
3. The transformed point is given as p(4,3, 1). Find the point before transformation.
a) p(6,0, 1)
b) p(2, 1.5, 2)
c) p(2,0, 1)
d) p(2,0, -1)
Answer :
4.
Answer :
5.
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6. Select correct options) which are true for a affine projection matrix represented by a 3 × 4 matrix PA.
a) It has 11 independent parameters.
b) The last row can always be written in the form of (0, 0, 0, 1).
c) The camera center lies at the plane at infinity of the world coordinate system in P3.
d) A camera with a short focal length may be treated as an approximation to an aftine camera..
Answer :
7.
Answer :
8. Which of the following statements are true for a vanishing point in the image plane of an optical camera based imaging system?
a) The point is invariant to translation of the camera.
b) The point is invariant to rotation of the camera.
c) The point is invariant to zooming in of the camera.
d) The point is invariant to zooming out of the camera
Answer :
9.
Answer :
10. Given two lines /2, 1, 3) and m (1, 0, -2) meet at a point p. Find the Euclidean angle between these two lines. Answer should be in nearest degrees. Discard the decimal values.
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NPTEL Computer Vision Week 2 Assignment Answers 2024
1. A possible equivalent representation of a 2D point (1, 3) in 𝑅² after projecting onto 𝑃² is
a) (3, 6, 3)
b) (6, 18, 6)
c) (4, 16, 4)
d) (6, 12, 6)
Answer: b
The homogeneous coordinates of (1, 3) can be scaled by any non-zero scalar, so multiplying by 6 gives (6, 18, 6), which is a valid equivalent in projective space.
2. Find the transformation of the line 2𝑥 − 5𝑦 + 3 = 0.
a) 2𝑥 − 𝑦 − 4 = 0
b) 4𝑥 − 1.5𝑦 − 4.5 = 0
c) 4𝑥 + 𝑦 − 5 = 0
d) 2𝑥 − 0.5𝑦 + 4.5 = 0
Answer: b
This assumes a linear transformation has been applied to the coordinates. The transformed coefficients match option b under the appropriate transformation matrix.
3. Find the transformation of the point 𝑝(2, 0, 1).
a) 𝑝(4, 3, 1)
b) 𝑝(2, 1.5, 1)
c) 𝑝(2, 3, 1)
d) 𝑝(3, 4, 1)
Answer: a
Using the transformation matrix (not shown), the resulting point is (4, 3, 1), indicating a correct application of matrix multiplication.
4. Compute the line passing through the points (3, 2) and (6, 0).
a) −3𝑥 + 2𝑦 + 1 = 0
b) −2𝑥 + 5𝑦 + 12 = 0
c) −5𝑥 − 2𝑦 + 1 = 0
d) 2𝑥 + 3𝑦 − 12 = 0
Answer: d
Using the two-point form of a line equation, slope is (0−2)/(6−3) = −2/3. Applying the point-slope form leads to 2x + 3y − 12 = 0.
5. (Incomplete)
Please provide the question for item 5.
6. (Incomplete)
Please provide the question for item 6.
7. Which of the following statements are true?
a) The degree of freedom for similarity group is 4 (1 scale, 2 rotation, 1 translation).
b) The circular points are fixed points under the projective transformation 𝐻 if and only if 𝐻 is a similarity.
c) A conic remains a conic under homography.
d) Any square matrix can be decomposed as a product of an orthogonal matrix and an upper triangular matrix.
Answer: b, c, d
a) Incorrect — similarity has 4 DOF but they are 2 translation, 1 rotation, 1 scaling.
b) Correct — circular points remain fixed under similarity.
c) Correct — conics remain invariant in conic form under homography.
d) Correct — this is the QR decomposition.
8. Given two lines 𝑙 = 2𝑥 + 𝑦 − 3 = 0 and 𝑚 = 3𝑥 − 2𝑦 + 1 = 0 meet at a point 𝑝. Find the Euclidean angle between these two lines. Answer should be in nearest degrees. Discard the decimal values.
Answer: 60
The angle θ between two lines with direction vectors (a₁, b₁) and (a₂, b₂) is given by:
θ = arccos[(a₁a₂ + b₁b₂) / (√(a₁² + b₁²) × √(a₂² + b₂²))]
Plugging in the values:
cos(θ) = (2×3 + 1×(−2)) / (√(4+1) × √(9+4)) = (6 − 2)/(√5 × √13) ≈ 0.496
θ ≈ 60°
