CBSE Class 12 Maths Previous Year Question Paper 2014
This is the CBSE Class 12 Maths Previous Year Question Paper from 2014, focusing on Dot & Cross Products of Two Vectors. The paper includes questions carrying 1 mark each, as seen in the provided excerpts. For instance, one question asks to find the magnitude of vector b given that vectors a and b are perpendicular, |a+b|=13, and |a|=5. Another question requires finding the angle between two unit vectors a and b when their sum a+b is also a unit vector. A third question asks for the projection of one vector onto another. Solving such previous year papers is crucial for students to understand the exam pattern, question types, and marking scheme, thereby enhancing their preparation and performance in the board examinations.
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Quick info
| Board | CBSE |
|---|---|
| Class | 12 |
| Subject | Maths |
| Session | 2014 |
| Language | English |
| Type | Previous Year Question Paper |
| Exam type | Board Exam |
Paper pattern
The paper includes 1-mark questions as part of the 2014 board examination.
Topics covered
Paper topics
- Vector Algebra
- Dot Product
- Cross Product
- Perpendicular Vectors
- Unit Vectors
- Vector Projection
Important topics
- Dot & Cross Products of Two Vectors
- Vector Projection
PDF preview
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Question paper text
Dot & Cross Products of Two Vectors
1 Marks Questions
1. If <math>\vec{a}</math> and <math>\vec{b}</math> are perpendicular vectors, <math>|\vec{a} + \vec{b}| = 13</math> and <math>|\vec{a}| = 5</math>, then find the value of <math>|\vec{b}|</math>. All India 2014
Given, <math>|\vec{a} + \vec{b}| = 13</math>, and <math>|\vec{a}| = 5</math>
Now,
<math>(\overrightarrow{a} + \overrightarrow{b}) \cdot (\overrightarrow{a} + \overrightarrow{b}) = \overrightarrow{a} \cdot \overrightarrow{a} + \overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{b} \cdot \overrightarrow{a} + \overrightarrow{b} \cdot \overrightarrow{b}</math>
<math>= |\vec{a}|^2 + 0 + 0 + |\vec{b}|^2</math>
<math>[\because \vec{a} \cdot \vec{a} = |\vec{a}|^2, \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = 0 \text{ as } \vec{a} \perp \vec{b}]</math> <math>\Rightarrow |\overrightarrow{a} + \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2</math> <math>\Rightarrow</math> <math>(13)^2 = (5)^2 + |\overrightarrow{b}|^2</math> <math>\Rightarrow</math> 169 = 25 + <math>|\vec{b}|^2 \Rightarrow</math> 169 - 25 = <math>|\vec{b}|^2</math> <math>\Rightarrow</math> <math display="block">144 = |\overrightarrow{b}|^2 \implies |\overrightarrow{b}| = 12</math> (1)
2. If <math>\overrightarrow{a}</math> and <math>\overrightarrow{b}</math> are two unit vectors such that <math>\vec{a} + \vec{b}</math> is also a unit vector, then find the angle between <math>\overrightarrow{a}</math> and <math>\overrightarrow{b}</math>. Delhi 2014
Given, <math>|\vec{a}| = 1, |\vec{b}| = 1</math> and <math>|\vec{a} + \vec{b}| = 1</math> Now, <math>|\overrightarrow{a} + \overrightarrow{b}|^2 = (\overrightarrow{a} + \overrightarrow{b}) \cdot (\overrightarrow{a} + \overrightarrow{b})</math> <math>\vec{a} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}</math> <math display="block">\Rightarrow |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2</math> <math display="block">[\because \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \text{ and } \vec{x} \cdot \vec{x} = |\vec{x}|^2]</math>
<math>\Rightarrow</math> <math>1=1+2\overrightarrow{a}\cdot\overrightarrow{b}+1</math> <math>2\vec{a}\cdot\vec{b} = -1</math> <math display="block">\Rightarrow |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta = -\frac{1}{2} [\because \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta]</math> <math>\Rightarrow</math> <math display="block">\cos \theta = -\frac{1}{2} \qquad [\because |\overrightarrow{a}| = |\overrightarrow{b}| = 1]</math>
<math>\Rightarrow</math> <math>\cos \theta = \cos \frac{2\pi}{3} \implies \theta = \frac{2\pi}{3}</math>
Hence, angle between <math>\vec{a}</math> and <math>\vec{b}</math> is <math>\frac{2\pi}{3}</math>. (1)
3. Find the projection of the vector <math>\hat{i} + 3\hat{j} + 7\hat{k}</math> on the vector <math>2\hat{i} - 3\hat{j} + 6\hat{k}</math>. Delhi 2014
Frequently asked questions
What is this document?
This is a CBSE Class 12 Maths Previous Year Question Paper from the 2014 board examinations.
What topics are covered?
This specific excerpt covers Dot & Cross Products of Two Vectors, including concepts like perpendicular vectors, unit vectors, and vector projection.
How does solving this paper help?
Solving previous year question papers helps students understand the board exam pattern, question difficulty, and marking scheme, leading to better preparation and improved scores.
What is the marking scheme for these questions?
The provided text indicates these are 1-mark questions from the 2014 CBSE Class 12 Maths paper.
Where can I find more questions?
This document contains a selection of questions from the 2014 paper. For comprehensive practice, refer to the complete previous year question papers.
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