Is this solution according to PECTAA syllabus?

Yes, these solutions follow the PECTAA / National Curriculum 2023 syllabus for Class 10 Mathematics.

Exercise 6.2 Solution — Chapter 6 Vectors in Plane | Class 10 Math

Q: How many questions are there in Chapter 6 Exercise 6.2?

Exercise 6.2 has 6 main questions covering vector addition, scalar multiplication, magnitude of combinations, and finding unknown components.

Q: What is vector addition?

Vector addition adds corresponding components: (x1 i-hat + y1 j-hat) + (x2 i-hat + y2 j-hat) = (x1+x2) i-hat + (y1+y2) j-hat.

Q: What is scalar multiplication of a vector?

Scalar multiplication multiplies a vector by a real number: k(x i-hat + y j-hat) = kx i-hat + ky j-hat.

Q: How do you find the magnitude of a vector combination?

First combine the vectors by addition and scalar multiplication, then use the formula |a| = sqrt(x squared + y squared).

Q: Are solved PDF notes available for Exercise 6.2?

Yes, a complete solved PDF for Exercise 6.2 is embedded on this page.

Q: Is Exercise 6.2 important for Class 10 board exams?

Yes, vector operations are frequently tested in board exams.

Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska

📌 Based on National Curriculum 2023 / PECTAA Syllabus

📖 What's Inside: This exercise covers vector addition, scalar multiplication, magnitude, and component equations. Perfect for Punjab Boards exam preparation.

⬇️ Download PDF (Exercise 6.2 Solved)

📚 Related Resources – Chapter 6: Vectors in Plane

🔢 Exercise 6.1 📐 Exercise 6.3 🔄 Review Exercise 6

Class 10 Math Chapter 6 Exercise 6.2 – Vector Operations: Complete Guide

Exercise 6.2 of Chapter 6 takes the foundational concepts from Exercise 6.1 and moves them into vector operations. This exercise is where students learn to combine vectors through addition, scale them using scalar multiplication, find the magnitude of vector combinations, and solve for unknown components in vector equations.

What You Will Learn

By working through Exercise 6.2, students master vector addition, scalar multiplication, combining both operations to simplify expressions, calculating the magnitude of combined vectors, and solving for unknown components by equating coefficients of î and ĵ.

Topics Covered in This Exercise

Vector addition – adding two or more vectors component-wise
Scalar multiplication – multiplying a vector by a real scalar
Combined operations – expressions like \(3\bar{a} + \frac{1}{2}\bar{b}\)
Magnitude of vector combinations – finding \(|\bar{a} - 2\bar{b} + \bar{c}|\)
Component equations – solving for \(x\) and \(y\) by equating î and ĵ coefficients

Why Exercise 6.2 Is Important

Exercise 6.2 is the operational core of Chapter 6. Board exam papers frequently feature questions that require students to add vectors, multiply by scalars, and compute magnitudes. Without these skills, students cannot successfully tackle Exercise 6.3 or the Review Exercise.

Punjab Board Preparation

Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this exercise. The combination of vector addition and scalar multiplication is one of the most common question types. The component-equation questions (like Q3 and Q6) are particularly popular with examiners.

Exam Tips for Vector Operations

Always group î and ĵ terms separately when simplifying.
When finding magnitude of a combination, first simplify the vector fully, then apply the magnitude formula.
For component equations, equate the coefficients of î and ĵ on both sides.
Double-check signs when subtracting vectors.

Common Mistakes Students Make

Forgetting to distribute a scalar to both components of a vector.
Adding the coefficients of î and ĵ together incorrectly.
Confusing \(|\bar{a} + \bar{b}|\) with \(|\bar{a}| + |\bar{b}|\) — these are not equal in general.
Not simplifying final answers, especially fractions and surds.

Muhammad Tayyab Subject Specialist Mathematics

MSc Mathematics · Govt Christian High School Daska, Sialkot, Punjab

Content reviewed against the PECTAA / National Curriculum 2023 syllabus for Class 10 Mathematics, applicable to all 10 BISE Punjab boards.

Last updated: August 5, 2026 Source: Punjab Curriculum & Textbook Board (PCTB)

1 Vector Addition & Scalar Multiplication

Given \(\bar{a} = 7\hat{i} - 3\hat{j}\) and \(\bar{b} = \hat{i} + 5\hat{j}\)

(i) \(\bar{a} + \bar{b}\)

\[ \begin{aligned} \bar{a} + \bar{b} &= (7\hat{i} - 3\hat{j}) + (\hat{i} + 5\hat{j}) \\ &= (7\hat{i} + \hat{i}) + (-3\hat{j} + 5\hat{j}) \\ &= 8\hat{i} + 2\hat{j} \end{aligned} \]

(ii) \(\bar{a} + 3\bar{b}\)

\[ \begin{aligned} \bar{a} + 3\bar{b} &= (7\hat{i} - 3\hat{j}) + 3(\hat{i} + 5\hat{j}) \\ &= 7\hat{i} - 3\hat{j} + 3\hat{i} + 15\hat{j} \\ &= 10\hat{i} + 12\hat{j} \end{aligned} \]

(iii) \(3\bar{a} + \frac{1}{2}\bar{b}\)

\[ \begin{aligned} 3\bar{a} + \frac{1}{2}\bar{b} &= 3(7\hat{i} - 3\hat{j}) + \frac{1}{2}(\hat{i} + 5\hat{j}) \\ &= 21\hat{i} - 9\hat{j} + \frac{1}{2}\hat{i} + \frac{5}{2}\hat{j} \\ &= \frac{43}{2}\hat{i} - \frac{13}{2}\hat{j} \end{aligned} \]

(iv) \(\bar{b} - \bar{a}\)

\[ \begin{aligned} \bar{b} - \bar{a} &= (\hat{i} + 5\hat{j}) - (7\hat{i} - 3\hat{j}) \\ &= \hat{i} + 5\hat{j} - 7\hat{i} + 3\hat{j} \\ &= -6\hat{i} + 8\hat{j} \end{aligned} \]

(v) \(4\bar{b} - 5\bar{a}\)

\[ \begin{aligned} 4\bar{b} - 5\bar{a} &= 4(\hat{i} + 5\hat{j}) - 5(7\hat{i} - 3\hat{j}) \\ &= 4\hat{i} + 20\hat{j} - 35\hat{i} + 15\hat{j} \\ &= -31\hat{i} + 35\hat{j} \end{aligned} \]

(vi) \(\frac{3}{2}\bar{a} - \bar{b}\)

\[ \begin{aligned} \frac{3}{2}\bar{a} - \bar{b} &= \frac{3}{2}(7\hat{i} - 3\hat{j}) - (\hat{i} + 5\hat{j}) \\ &= \frac{21}{2}\hat{i} - \frac{9}{2}\hat{j} - \hat{i} - 5\hat{j} \\ &= \frac{19}{2}\hat{i} - \frac{19}{2}\hat{j} \end{aligned} \]

2 Magnitude of Vector Combinations

Given \(\bar{a} = 6\hat{i} - \hat{j}\), \(\bar{b} = \hat{i} + 5\hat{j}\), \(\bar{c} = 3\hat{i} + 5\hat{j}\)

(i) \(\bar{b} - \bar{c}\)

\[ \begin{aligned} \bar{b} - \bar{c} &= (\hat{i} + 5\hat{j}) - (3\hat{i} + 5\hat{j}) \\ &= \hat{i} - 3\hat{i} + 5\hat{j} - 5\hat{j} \\ &= -2\hat{i} + 0\hat{j} \\ |\bar{b} - \bar{c}| &= \sqrt{(-2)^2 + 0^2} = 2 \end{aligned} \]

(ii) \(\bar{a} - 2\bar{b} + \bar{c}\)

\[ \begin{aligned} \bar{a} - 2\bar{b} + \bar{c} &= (6\hat{i} - \hat{j}) - 2(\hat{i} + 5\hat{j}) + (3\hat{i} + 5\hat{j}) \\ &= 6\hat{i} - \hat{j} - 2\hat{i} - 10\hat{j} + 3\hat{i} + 5\hat{j} \\ &= (6-2+3)\hat{i} + (-1-10+5)\hat{j} \\ &= 7\hat{i} - 6\hat{j} \\ |\bar{a} - 2\bar{b} + \bar{c}| &= \sqrt{7^2 + (-6)^2} = \sqrt{85} \end{aligned} \]

(iii) \(\bar{c} - \bar{b} - \bar{a}\)

\[ \begin{aligned} \bar{c} - \bar{b} - \bar{a} &= (3\hat{i} + 5\hat{j}) - (\hat{i} + 5\hat{j}) - (6\hat{i} - \hat{j}) \\ &= 3\hat{i} + 5\hat{j} - \hat{i} - 5\hat{j} - 6\hat{i} + \hat{j} \\ &= (3-1-6)\hat{i} + (5-5+1)\hat{j} \\ &= -4\hat{i} + \hat{j} \\ |\bar{c} - \bar{b} - \bar{a}| &= \sqrt{(-4)^2 + 1^2} = \sqrt{17} \end{aligned} \]

3 Find x and y from Vector Equations

(i) \((x\hat{i} + y\hat{j}) + (2\hat{i} + 3\hat{j}) = 7\hat{i} + 6\hat{j}\)

\[ \begin{aligned} (x\hat{i} + y\hat{j}) + (2\hat{i} + 3\hat{j}) &= 7\hat{i} + 6\hat{j} \\ x\hat{i} + y\hat{j} + 2\hat{i} + 3\hat{j} &= 7\hat{i} + 6\hat{j} \\ (x+2)\hat{i} + (y+3)\hat{j} &= 7\hat{i} + 6\hat{j} \end{aligned} \]

Equating components:

\[ \begin{aligned} x+2 &= 7 \implies x = 5 \\ y+3 &= 6 \implies y = 3 \end{aligned} \]

(ii) \((x\hat{i} - 5\hat{j}) + (3\hat{i} + 5\hat{j}) = -8\hat{i} + y\hat{j}\)

\[ \begin{aligned} (x\hat{i} - 5\hat{j}) + (3\hat{i} + 5\hat{j}) &= -8\hat{i} + y\hat{j} \\ x\hat{i} + 3\hat{i} - 5\hat{j} + 5\hat{j} &= -8\hat{i} + y\hat{j} \\ (x+3)\hat{i} + 0\hat{j} &= -8\hat{i} + y\hat{j} \end{aligned} \]

Equating components:

\[ \begin{aligned} x+3 &= -8 \implies x = -11 \\ 0 &= y \implies y = 0 \end{aligned} \]

(iii) \((y\hat{i} + 3\hat{j}) + (-5\hat{i} + 2x\hat{j}) = 9\hat{i} + 7\hat{j}\)

\[ \begin{aligned} (y\hat{i} + 3\hat{j}) + (-5\hat{i} + 2x\hat{j}) &= 9\hat{i} + 7\hat{j} \\ y\hat{i} - 5\hat{i} + 3\hat{j} + 2x\hat{j} &= 9\hat{i} + 7\hat{j} \\ (y-5)\hat{i} + (3+2x)\hat{j} &= 9\hat{i} + 7\hat{j} \end{aligned} \]

Equating components:

\[ \begin{aligned} y-5 &= 9 \implies y = 14 \\ 3+2x &= 7 \implies 2x = 4 \implies x = 2 \end{aligned} \]

4 Find Magnitude of \(\bar{b}\)

Given \(\bar{a} = \hat{i} + 3\hat{j}\), \(\bar{c} = 2\hat{i} + \hat{j}\) and \(\bar{a} + 2\bar{b} = \bar{c}\)

\[ \begin{aligned} \bar{a} + 2\bar{b} &= \bar{c} \\ (\hat{i} + 3\hat{j}) + 2\bar{b} &= 2\hat{i} + \hat{j} \\ 2\bar{b} &= (2\hat{i} + \hat{j}) - (\hat{i} + 3\hat{j}) \\ 2\bar{b} &= \hat{i} - 2\hat{j} \\ \bar{b} &= \frac{1}{2}\hat{i} - \hat{j} \end{aligned} \]

Now, magnitude:

\[ |\bar{b}| = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \]

5 Vector Components of \(\bar{a} + 5\bar{b}\)

Given \(\bar{a} = -2\hat{i} + 7\hat{j}\), \(\bar{b} = 3\hat{i} - 5\hat{j}\)

\[ \begin{aligned} \bar{a} + 5\bar{b} &= (-2\hat{i} + 7\hat{j}) + 5(3\hat{i} - 5\hat{j}) \\ &= -2\hat{i} + 7\hat{j} + 15\hat{i} - 25\hat{j} \\ &= 13\hat{i} - 18\hat{j} \end{aligned} \]

6 Find Values of m and n

Given \(5\hat{i} - 3\hat{j} = m(\hat{i} - 10\hat{j}) + n(4\hat{i} - 3\hat{j})\)

\[ \begin{aligned} 5\hat{i} - 3\hat{j} &= m(\hat{i} - 10\hat{j}) + n(4\hat{i} - 3\hat{j}) \\ &= m\hat{i} - 10m\hat{j} + 4n\hat{i} - 3n\hat{j} \\ &= (m+4n)\hat{i} + (-10m-3n)\hat{j} \end{aligned} \]

Equating components:

\[ \begin{aligned} m + 4n &= 5 \quad \text{(i)} \\ -10m - 3n &= -3 \quad \text{(ii)} \end{aligned} \]

From (i): \(m = 5 - 4n\). Substituting in (ii):

\[ \begin{aligned} -10(5 - 4n) - 3n &= -3 \\ -50 + 40n - 3n &= -3 \\ 37n &= 47 \\ n &= \frac{47}{37} \end{aligned} \]

Putting \(n\) in equation (i):

\[ \begin{aligned} m + 4\left(\frac{47}{37}\right) &= 5 \\ m &= 5 - \frac{188}{37} = \frac{185 - 188}{37} = -\frac{3}{37} \end{aligned} \]

📈 Key Concepts & Quick Revision

Vector Addition: Add corresponding components: \((x_1\hat{i} + y_1\hat{j}) + (x_2\hat{i} + y_2\hat{j}) = (x_1+x_2)\hat{i} + (y_1+y_2)\hat{j}\)
Scalar Multiplication: \(k(x\hat{i} + y\hat{j}) = kx\hat{i} + ky\hat{j}\)
Magnitude: \(|\bar{a}| = \sqrt{x^2 + y^2}\)
Vector Equality: Two vectors are equal if their corresponding components are equal.

🎯 Important Definitions

Vector: a quantity with both magnitude and direction.
Scalar: a real number used to scale a vector.
Resultant Vector: the sum of two or more vectors.

📝 Important MCQs for Practice

1. If \(\bar{a} = 2\hat{i} + 3\hat{j}\) and \(\bar{b} = 4\hat{i} - \hat{j}\), then \(\bar{a} + \bar{b}\) equals:

(a) \(6\hat{i} + 2\hat{j}\) (b) \(6\hat{i} - 2\hat{j}\) (c) \(-2\hat{i} + 4\hat{j}\) (d) none — Answer: (a)

2. If \(\bar{a} = 3\hat{i} - 4\hat{j}\), then \(|\bar{a}|\) equals:

(a) 5 (b) 7 (c) 12 (d) 25 — Answer: (a)

3. If \(\bar{a} = 2\hat{i} - 3\hat{j}\), then \(2\bar{a}\) equals:

(a) \(4\hat{i} - 6\hat{j}\) (b) \(4\hat{i} + 6\hat{j}\) (c) \(-4\hat{i} + 6\hat{j}\) (d) none — Answer: (a)

🏆 Board Exam Tips & Strategy

Show all steps when simplifying vector expressions — examiners award marks for intermediate steps.
For component equations, write out the equation clearly before equating coefficients.
Leave answers in exact surd form unless a decimal is required.
Practice simultaneous equations for component problems — they appear frequently.

❓ Frequently Asked Questions

What is taught in Exercise 6.2 of Class 10 Math Chapter 6?

Exercise 6.2 covers vector addition, scalar multiplication, magnitude of vector combinations, and solving component equations.

How many questions are there in Chapter 6 Exercise 6.2?

Exercise 6.2 has 6 main questions covering vector operations and component equations.

Is this solution according to the PECTAA syllabus?

Yes, these solutions are prepared according to the PECTAA / National Curriculum 2023 syllabus.

Is this Exercise 6.2 solution valid for all Punjab Boards?

Yes, the content follows the unified Punjab textbook and is applicable to all 10 BISE Punjab boards.

What is vector addition?

Vector addition is the operation of adding two vectors component-wise: \((x_1\hat{i} + y_1\hat{j}) + (x_2\hat{i} + y_2\hat{j}) = (x_1+x_2)\hat{i} + (y_1+y_2)\hat{j}\).

What is scalar multiplication of a vector?

Scalar multiplication is multiplying a vector by a real number: \(k(x\hat{i} + y\hat{j}) = kx\hat{i} + ky\hat{j}\).

How do you find the magnitude of a vector combination?

First combine the vectors, then use the formula \(|\bar{a}| = \sqrt{x^2 + y^2}\).

Are solved PDF notes available for Exercise 6.2?

Yes, a complete solved PDF is embedded on this page and available for free download.

Can I download the Chapter 6 Exercise 6.2 solution as a PDF?

Yes, use the Download PDF button on this page to save the complete solved Exercise 6.2 notes.

Is Exercise 6.2 important for Class 10 board exams?

Yes, vector operations are frequently tested, and this exercise builds essential skills for later exercises.

Who prepared these Class 10 Math Chapter 6 notes?

These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska.

📘 Exercise 6.1 📘 Exercise 6.3 📘 Review Exercise 6 📗 Chapter 6 Hub – Vectors in Plane 📚 Class 10 Mathematics – All Chapters 🏠 Hira Science Academy – Home

Class 10 Math Chapter 6 Exercise 6.2 – Vector Operations: Complete Guide

What You Will Learn

Topics Covered in This Exercise

Why Exercise 6.2 Is Important

Punjab Board Preparation

Exam Tips for Vector Operations

Common Mistakes Students Make

📈 Key Concepts & Quick Revision

🎯 Important Definitions

📝 Important MCQs for Practice

🏆 Board Exam Tips & Strategy

❓ Frequently Asked Questions

🔗 Related Resources