· ×
📐 Chapter 9: Product of Vectors | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Dot Product, Cross Product & Applications
✅ Updated according to latest syllabus of 2080 | Complete notes with MCQs and practice questions
🔷 Product of Vectors — This chapter covers scalar product (dot product) and vector product (cross product) of vectors. The notes have been updated according to the latest syllabus of 2080. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📚 Exercises
📖 Introduction
Product of Vectors: The multiplication operation between two vectors can be performed in two ways:
• Scalar Product (Dot Product): Result is a scalar quantity.
• Vector Product (Cross Product): Result is a vector quantity.
• Scalar Product (Dot Product): Result is a scalar quantity.
• Vector Product (Cross Product): Result is a vector quantity.
⚫ Scalar Product (Dot Product)
Definition: The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \] where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
Properties:
• \(\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\) (Commutative)
• \(\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}\) (Distributive)
• \(\vec{A} \cdot \vec{A} = |\vec{A}|^2\)
• \(\vec{A} \cdot \vec{B} = 0\) if \(\vec{A} \perp \vec{B}\)
• In component form: \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\), \(\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\)
\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \] where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
Properties:
• \(\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\) (Commutative)
• \(\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}\) (Distributive)
• \(\vec{A} \cdot \vec{A} = |\vec{A}|^2\)
• \(\vec{A} \cdot \vec{B} = 0\) if \(\vec{A} \perp \vec{B}\)
• In component form: \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\), \(\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\)
\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]
🔴 Vector Product (Cross Product)
Definition: The vector product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as:
\[ \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin \theta \, \hat{n} \] where \(\hat{n}\) is the unit vector perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\).
Properties:
• \(\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})\) (Anti-commutative)
• \(\vec{A} \times \vec{A} = 0\)
• \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\)
• In component form: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
\[ \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin \theta \, \hat{n} \] where \(\hat{n}\) is the unit vector perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\).
Properties:
• \(\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})\) (Anti-commutative)
• \(\vec{A} \times \vec{A} = 0\)
• \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\)
• In component form: \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
📐 Angle Between Two Vectors
Formula: The angle \(\theta\) between two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Special Cases:
• \(\theta = 0^\circ\) (Parallel vectors): \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}|\)
• \(\theta = 90^\circ\) (Perpendicular vectors): \(\vec{A} \cdot \vec{B} = 0\)
• \(\theta = 180^\circ\) (Anti-parallel): \(\vec{A} \cdot \vec{B} = -|\vec{A}| |\vec{B}|\)
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Special Cases:
• \(\theta = 0^\circ\) (Parallel vectors): \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}|\)
• \(\theta = 90^\circ\) (Perpendicular vectors): \(\vec{A} \cdot \vec{B} = 0\)
• \(\theta = 180^\circ\) (Anti-parallel): \(\vec{A} \cdot \vec{B} = -|\vec{A}| |\vec{B}|\)
📝 Solved Problems
Q1 If \(\vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k}\) and \(\vec{B} = \hat{i} – 2\hat{j} + \hat{k}\), find \(\vec{A} \cdot \vec{B}\).
Solution:
\(\vec{A} \cdot \vec{B} = (2)(1) + (3)(-2) + (4)(1) = 2 – 6 + 4 = 0\)
\(\vec{A} \cdot \vec{B} = (2)(1) + (3)(-2) + (4)(1) = 2 – 6 + 4 = 0\)
✅ \(\vec{A} \cdot \vec{B} = 0\) (Vectors are perpendicular)
Q2 Find the angle between \(\vec{A} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} – \hat{j} + \hat{k}\).
Solution:
\(|\vec{A}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}\)
\(|\vec{B}| = \sqrt{1^2+(-1)^2+1^2} = \sqrt{3}\)
\(\vec{A} \cdot \vec{B} = 1(1) + 1(-1) + 1(1) = 1 – 1 + 1 = 1\)
\(\cos \theta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}\) ⇒ \(\theta = \cos^{-1}(\frac{1}{3})\)
\(|\vec{A}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}\)
\(|\vec{B}| = \sqrt{1^2+(-1)^2+1^2} = \sqrt{3}\)
\(\vec{A} \cdot \vec{B} = 1(1) + 1(-1) + 1(1) = 1 – 1 + 1 = 1\)
\(\cos \theta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}\) ⇒ \(\theta = \cos^{-1}(\frac{1}{3})\)
✅ \(\theta = \cos^{-1}(\frac{1}{3})\)
Q3 If \(\vec{A} = 2\hat{i} + 3\hat{j} – \hat{k}\) and \(\vec{B} = \hat{i} – 2\hat{j} + \hat{k}\), find \(\vec{A} \times \vec{B}\).
Solution:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(3\cdot1 – (-1)(-2)) – \hat{j}(2\cdot1 – (-1)(1)) + \hat{k}(2\cdot(-2) – 3\cdot1) \] \[ = \hat{i}(3 – 2) – \hat{j}(2 + 1) + \hat{k}(-4 – 3) = \hat{i} – 3\hat{j} – 7\hat{k} \]
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(3\cdot1 – (-1)(-2)) – \hat{j}(2\cdot1 – (-1)(1)) + \hat{k}(2\cdot(-2) – 3\cdot1) \] \[ = \hat{i}(3 – 2) – \hat{j}(2 + 1) + \hat{k}(-4 – 3) = \hat{i} – 3\hat{j} – 7\hat{k} \]
✅ \(\vec{A} \times \vec{B} = \hat{i} – 3\hat{j} – 7\hat{k}\)
Q4 Find the area of the parallelogram whose adjacent sides are \(\vec{A} = 2\hat{i} + 3\hat{j}\) and \(\vec{B} = \hat{i} + 4\hat{j}\).
Solution:
Area = \(|\vec{A} \times \vec{B}|\)
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & 4 & 0 \end{vmatrix} = \hat{k}(2\cdot4 – 3\cdot1) = 5\hat{k}\)
\(|\vec{A} \times \vec{B}| = 5\)
Area = \(|\vec{A} \times \vec{B}|\)
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & 4 & 0 \end{vmatrix} = \hat{k}(2\cdot4 – 3\cdot1) = 5\hat{k}\)
\(|\vec{A} \times \vec{B}| = 5\)
✅ Area = 5 square units
Q5 Find the unit vector perpendicular to both \(\vec{A} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} – \hat{j} + \hat{k}\).
Solution:
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1\cdot1 – 1\cdot(-1)) – \hat{j}(1\cdot1 – 1\cdot1) + \hat{k}(1\cdot(-1) – 1\cdot1)\)
\(= \hat{i}(1 + 1) – \hat{j}(1 – 1) + \hat{k}(-1 – 1) = 2\hat{i} + 0\hat{j} – 2\hat{k}\)
\(|\vec{A} \times \vec{B}| = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}\)
Unit vector = \(\frac{2\hat{i} – 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i} – \hat{k}}{\sqrt{2}}\)
\(\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1\cdot1 – 1\cdot(-1)) – \hat{j}(1\cdot1 – 1\cdot1) + \hat{k}(1\cdot(-1) – 1\cdot1)\)
\(= \hat{i}(1 + 1) – \hat{j}(1 – 1) + \hat{k}(-1 – 1) = 2\hat{i} + 0\hat{j} – 2\hat{k}\)
\(|\vec{A} \times \vec{B}| = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}\)
Unit vector = \(\frac{2\hat{i} – 2\hat{k}}{2\sqrt{2}} = \frac{\hat{i} – \hat{k}}{\sqrt{2}}\)
✅ \(\hat{n} = \frac{\hat{i} – \hat{k}}{\sqrt{2}}\)
✏️ Practice Questions for Students
P1 If \(\vec{A} = 3\hat{i} + 4\hat{j} – 2\hat{k}\) and \(\vec{B} = 2\hat{i} – \hat{j} + 5\hat{k}\), find \(\vec{A} \cdot \vec{B}\).
✅ Answer: \(3(2) + 4(-1) + (-2)(5) = 6 – 4 – 10 = -8\)
P2 Find the angle between \(\vec{A} = 2\hat{i} + 2\hat{j} – \hat{k}\) and \(\vec{B} = 6\hat{i} – 3\hat{j} + 2\hat{k}\).
✅ Answer: \(\cos \theta = \frac{4}{21}\), \(\theta = \cos^{-1}(\frac{4}{21})\)
P3 If \(\vec{A} = 2\hat{i} + 3\hat{j} – \hat{k}\) and \(\vec{B} = \hat{i} + \hat{j} – 2\hat{k}\), find \(|\vec{A} \times \vec{B}|\).
✅ Answer: \(|\vec{A} \times \vec{B}| = \sqrt{(-5)^2 + 3^2 + (-1)^2} = \sqrt{25+9+1} = \sqrt{35}\)
P4 Find the area of the triangle with vertices A(1,2,3), B(2,3,4), C(3,4,5).
✅ Answer: Vectors are collinear, area = 0
P5 For what value of \(\lambda\) are the vectors \(\vec{A} = 2\hat{i} + \lambda\hat{j} + \hat{k}\) and \(\vec{B} = 4\hat{i} – 2\hat{j} + 2\hat{k}\) perpendicular?
✅ Answer: \(\vec{A} \cdot \vec{B} = 8 – 2\lambda + 2 = 0\) ⇒ \(10 – 2\lambda = 0\) ⇒ \(\lambda = 5\)
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 The dot product of two perpendicular vectors is:
A) Maximum B) Minimum C) Zero D) Unity
✅ Answer: C) Zero
MCQ 2 If \(\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|\), then the angle between \(\vec{A}\) and \(\vec{B}\) is:
A) 0° B) 30° C) 45° D) 60°
Solution: \(|\vec{A}||\vec{B}|\cos\theta = |\vec{A}||\vec{B}|\sin\theta\) ⇒ \(\tan\theta = 1\) ⇒ \(\theta = 45°\)
✅ Answer: C) 45°
MCQ 3 \(\hat{i} \cdot (\hat{j} \times \hat{k})\) is equal to:
A) 0 B) 1 C) -1 D) \(\hat{i}\)
\(\hat{j} \times \hat{k} = \hat{i}\), then \(\hat{i} \cdot \hat{i} = 1\)
✅ Answer: B) 1
MCQ 4 The cross product of two parallel vectors is:
A) Maximum B) Zero vector C) Unit vector D) Scalar
✅ Answer: B) Zero vector
MCQ 5 If \(|\vec{A} \times \vec{B}| = \vec{A} \cdot \vec{B}\), then the angle between \(\vec{A}\) and \(\vec{B}\) is:
A) 0° B) 30° C) 45° D) 60°
✅ Answer: C) 45°
MCQ 6 The value of \(\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})\) is:
A) 0 B) 1 C) 2 D) 3
Each term = 1, sum = 3
✅ Answer: D) 3
MCQ 7 Two vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A} + \vec{B}| = |\vec{A} – \vec{B}|\). The angle between them is:
A) 0° B) 60° C) 90° D) 180°
Squaring both sides: \(|\vec{A}|^2 + |\vec{B}|^2 + 2\vec{A}\cdot\vec{B} = |\vec{A}|^2 + |\vec{B}|^2 – 2\vec{A}\cdot\vec{B}\) ⇒ \(4\vec{A}\cdot\vec{B}=0\) ⇒ \(\vec{A}\cdot\vec{B}=0\) ⇒ \(\theta = 90°\)
✅ Answer: C) 90°
MCQ 8 The magnitude of the cross product of two unit vectors \(\hat{i}\) and \(\hat{j}\) is:
A) 0 B) 1 C) \(\sqrt{2}\) D) 2
✅ Answer: B) 1
📚 More Solved Problems
Q6 Prove that \((\vec{A} + \vec{B}) \cdot (\vec{A} – \vec{B}) = |\vec{A}|^2 – |\vec{B}|^2\).
LHS = \(\vec{A}\cdot\vec{A} – \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} – \vec{B}\cdot\vec{B} = |\vec{A}|^2 – |\vec{B}|^2\) (since \(\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\))
✅ Hence proved
Q7 Find the projection of \(\vec{A} = 2\hat{i} + 3\hat{j} – \hat{k}\) on \(\vec{B} = \hat{i} – 2\hat{j} + 2\hat{k}\).
Projection = \(\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = \frac{2(1) + 3(-2) + (-1)(2)}{\sqrt{1 + 4 + 4}} = \frac{2 – 6 – 2}{3} = \frac{-6}{3} = -2\)
✅ Projection = \(-2\)
Q8 Find the work done by a force \(\vec{F} = 2\hat{i} + 3\hat{j} – \hat{k}\) when it displaces a body from A(1,2,3) to B(4,5,6).
Work = \(\vec{F} \cdot \vec{d}\) where \(\vec{d} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}\)
Work = \(2(3) + 3(3) + (-1)(3) = 6 + 9 – 3 = 12\) units
Work = \(2(3) + 3(3) + (-1)(3) = 6 + 9 – 3 = 12\) units
✅ Work = 12 units
⚠️ Please do not share this PDF on any website or social platform without permission.
This PDF provides the solutions of every question from the 1st exercise of class 12 product of vectors chapter. If you want the solutions of other exercises then you can select the exercise from the button given above.
This PDF provides the solutions of every question from the 1st exercise of class 12 product of vectors chapter. If you want the solutions of other exercises then you can select the exercise from the button given above.
📚 More Read
Class 12 Nepali Important Questions
NEB Class 12 English Important Questions
NEB Class 12 Biology Important Questions
NEB Class 12 Chemistry Important Questions
📌 Key Formulas Summary:
• Dot Product: \(\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3\)
• Cross Product: \(\vec{A} \times \vec{B} = |\vec{A}||\vec{B}|\sin\theta \, \hat{n}\)
• Angle between vectors: \(\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\)
• Area of parallelogram: \(|\vec{A} \times \vec{B}|\)
• Area of triangle: \(\frac{1}{2}|\vec{A} \times \vec{B}|\)
• Unit vector perpendicular to both: \(\hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\)
• Dot Product: \(\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3\)
• Cross Product: \(\vec{A} \times \vec{B} = |\vec{A}||\vec{B}|\sin\theta \, \hat{n}\)
• Angle between vectors: \(\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}\)
• Area of parallelogram: \(|\vec{A} \times \vec{B}|\)
• Area of triangle: \(\frac{1}{2}|\vec{A} \times \vec{B}|\)
• Unit vector perpendicular to both: \(\hat{n} = \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\)