Mathematics 2022 Paper I 50 marks Solve

Q3

(a) Let the set P = $\left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} \middle| \begin{array}{l} x - y - z = 0 \text{ and} \\ 2x - y + z = 0 \end{array} \right\}$ be the collection of vectors of a vector space $\mathbb{R}^3(\mathbb{R})$. Then (i) prove that P is a subspace of $\mathbb{R}^3$. (ii) find a basis and dimension of P. 10+10 (b) Use double integration to calculate the area common to the circle $x^2 + y^2 = 4$ and the parabola $y^2 = 3x$. 15 (c) Find the equation of the sphere of smallest possible radius which touches the straight lines : $\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$. 15

हिंदी में प्रश्न पढ़ें

(a) माना समुच्चय P = $\left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} \middle| \begin{array}{l} x - y - z = 0 \text{ तथा} \\ 2x - y + z = 0 \end{array} \right\}$ सदिश समष्टि $\mathbb{R}^3(\mathbb{R})$ के सदिशों का एक समुह है । तब (i) सिद्ध कीजिए कि P, $\mathbb{R}^3$ की एक उपसमष्टि है । (ii) P का एक आधार तथा विमा ज्ञात कीजिए । 10+10 (b) दिशा: समाकलन का उपयोग करके, वृत्त $x^2 + y^2 = 4$ तथा परवलय $y^2 = 3x$ के उभयनिष्ठ क्षेत्रफल का परिकलन कीजिए । 15 (c) लघुतम संभाव्य त्रिज्या के गोले का समीकरण ज्ञात कीजिए जो सरल रेखाओं : $\frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1}$ तथा $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ को स्पर्श करता है । 15

Directive word: Solve

This question asks you to solve. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

Solve all three parts systematically, allocating approximately 40% time to part (a) (20 marks), 30% to part (b) (15 marks), and 30% to part (c) (15 marks). Begin with clear statement of what is given and what is to be found for each sub-part, proceed with rigorous mathematical working, and conclude with boxed final answers for dimension, area, and sphere equation respectively.

Key points expected

  • For (a)(i): Verify P contains zero vector, is closed under vector addition, and closed under scalar multiplication using the two defining linear equations
  • For (a)(ii): Solve the homogeneous system to find parametric form, extract basis vectors, and state dimension of P (expected: dim(P) = 1)
  • For (b): Identify intersection points of circle x²+y²=4 and parabola y²=3x, set up correct double integral with proper limits (x from 0 to 1, y from -√(3x) to √(3x) plus appropriate circular segment), and evaluate to get common area
  • For (c): Find shortest distance between two skew lines using the formula involving cross product of direction vectors, determine midpoint of common perpendicular as sphere center, and write equation with this radius
  • For (c): Verify lines are indeed skew by checking (a₂-a₁)·(b₁×b₂) ≠ 0, then apply standard formula for minimum distance between skew lines

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10Correctly identifies P as solution space of homogeneous system for (a); accurately finds intersection points (1,±√3) for (b); verifies lines are skew and identifies all direction/position vectors for (c)Minor errors in setting up one part, such as wrong intersection points or partially correct vector identificationFundamental setup errors: treats P as non-homogeneous, wrong region for integration, or assumes lines are parallel/intersecting
Method choice20%10Uses subspace test/row reduction for (a); employs polar coordinates or appropriate Cartesian splitting for (b); applies shortest distance formula for skew lines for (c)Correct but inefficient methods, such as brute force integration without exploiting symmetry or lengthy alternative to standard formulasInappropriate methods: attempts basis without solving system, uses single integral for (b), or treats skew lines as coplanar
Computation accuracy20%10Exact arithmetic: dimension = 1, basis vector proportional to (2,3,1); area = (4π/3 + √3/2) or equivalent exact form; sphere center and radius computed precisely with exact coordinatesCorrect method with minor arithmetic slips leading to slightly wrong final values, or correct final answers with some intermediate steps omittedMajor computational errors: wrong determinant values, incorrect integration results, or algebraic mistakes in solving simultaneous equations
Step justification20%10Explicitly states subspace axioms being verified; shows Jacobian or area element justification; proves why shortest distance gives minimum radius sphere with clear geometric reasoningSome logical gaps in justification, such as asserting without proof that intersection of subspaces is subspace, or assuming without verification that common perpendicular existsMissing crucial justifications: no verification of closure properties, no explanation of integral limits, or unsupported claims about minimum radius
Final answer & units20%10Clear boxed answers: basis {(2,3,1)} or equivalent, dim(P)=1; area in exact simplified form with square units; sphere equation in standard form (x-a)²+(y-b)²+(z-c)²=r² with all values explicitly statedCorrect answers but poorly formatted, missing units for area, or sphere equation left in expanded form without identifying center and radius clearlyMissing final answers, wrong form of presentation (e.g., parametric instead of Cartesian for sphere), or numerical approximations where exact forms required

Practice this exact question

Write your answer, then get a detailed evaluation from our AI trained on UPSC's answer-writing standards. Free first evaluation — no signup needed to start.

Evaluate my answer →

More from Mathematics 2022 Paper I