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
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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
| Dimension | Weight | Max marks | Excellent | Average | Poor |
|---|---|---|---|---|---|
| Setup correctness | 20% | 10 | Correctly 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 identification | Fundamental setup errors: treats P as non-homogeneous, wrong region for integration, or assumes lines are parallel/intersecting |
| Method choice | 20% | 10 | Uses 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 formulas | Inappropriate methods: attempts basis without solving system, uses single integral for (b), or treats skew lines as coplanar |
| Computation accuracy | 20% | 10 | Exact 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 coordinates | Correct method with minor arithmetic slips leading to slightly wrong final values, or correct final answers with some intermediate steps omitted | Major computational errors: wrong determinant values, incorrect integration results, or algebraic mistakes in solving simultaneous equations |
| Step justification | 20% | 10 | Explicitly states subspace axioms being verified; shows Jacobian or area element justification; proves why shortest distance gives minimum radius sphere with clear geometric reasoning | Some logical gaps in justification, such as asserting without proof that intersection of subspaces is subspace, or assuming without verification that common perpendicular exists | Missing crucial justifications: no verification of closure properties, no explanation of integral limits, or unsupported claims about minimum radius |
| Final answer & units | 20% | 10 | Clear 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 stated | Correct answers but poorly formatted, missing units for area, or sphere equation left in expanded form without identifying center and radius clearly | Missing final answers, wrong form of presentation (e.g., parametric instead of Cartesian for sphere), or numerical approximations where exact forms required |
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