Miscellaneous Exercise answers: Introduction to Three Dimensional Geometry

Class 11 Maths4 questions

Miscellaneous Exercise

4 questions · page 215 of the book

Question 1

“Three vertices of a parallelogram ABCD are A(3, –1, 2), B (1, 2, –4) and C (–1, 1, 2). Find … the fourth vertex.” · p. 215

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  1. In a parallelogram, the diagonals bisect each other.
  2. The midpoint of diagonal AC = the midpoint of diagonal BD
  3. Midpoint of AC = ((3 − 1)/2, (−1 + 1)/2, (2 + 2)/2) = (1, 0, 2)
  4. Let D = (x, y, z). Then midpoint of BD = ((1 + x)/2, (2 + y)/2, (−4 + z)/2)
  5. Setting them equal: (1 + x)/2 = 1 ⟹ x = 1
  6. (2 + y)/2 = 0 ⟹ y = −2
  7. (−4 + z)/2 = 2 ⟹ z = 8
  8. Therefore, D = (1, −2, 8)

Answer(1, −2, 8)

Watch this explained “What settles a parallelogram”, 11:50 into Applying the right-triangle rule twice to get out of the plane

Question 2

“Find the lengths of the medians of the triangle with vertices …” · p. 215

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  1. Given: A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
  2. The median from a vertex goes to the midpoint of the opposite side.
  3. Midpoint of BC = (3, 2, 0). Median from A = √((3−0)² + (2−0)² + (0−6)²) = √49 = 7.
  4. Midpoint of AC = (3, 0, 3). Median from B = √((3−0)² + (0−4)² + (3−0)²) = √34.
  5. Midpoint of AB = (0, 2, 3). Median from C = √((0−6)² + (2−0)² + (3−0)²) = √49 = 7.

AnswerMedian from A = 7, median from B = √34, median from C = 7

Watch the lesson Applying the right-triangle rule twice to get out of the plane

Question 3

“If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R(8, 14, 2c) …” · p. 215

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  1. The centroid's coordinates are the average of the vertices' coordinates.
  2. x: (2a − 4 + 8)/3 = 0, so 2a + 4 = 0, giving a = −2.
  3. y: (2 + 3b + 14)/3 = 0, so 3b + 16 = 0, giving b = −16/3.
  4. z: (6 − 10 + 2c)/3 = 0, so 2c − 4 = 0, giving c = 2.

Answera = −2, b = −16/3, c = 2

Question 4

“… find the equation of the set of points P such that PA² + PB² = k², where k is a constant.” · p. 215

Open NCERT p. 215Matches NCERT’s answer

  1. Let P be (x, y, z).
  2. PA² = (x − 3)² + (y − 4)² + (z − 5)² = x² + y² + z² − 6x − 8y − 10z + 50.
  3. PB² = (x + 1)² + (y − 3)² + (z + 7)² = x² + y² + z² + 2x − 6y + 14z + 59.
  4. PA² + PB² = 2x² + 2y² + 2z² − 4x − 14y + 4z + 109.
  5. Set this equal to k²: 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = k², i.e. 2x² + 2y² + 2z² − 4x − 14y + 4z = k² − 109.

Answer2x² + 2y² + 2z² − 4x − 14y + 4z = k² − 109

Watch this explained “Conditions, equations, and a gap”, 13:00 into Applying the right-triangle rule twice to get out of the plane

Every question here was solved twice, separately, by two different AI models, and each answer was put back into the question by a computer program to check it. Where the two disagreed, a stronger model solved it again and the computer check had to pass on its answer. A question about reasoning rather than a number is shown as “one way to think about it”, and anything not yet proven says so instead of guessing. Each answer links to the moment in the video that teaches it.

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