Vector Algebra — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Vector Algebra" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Vector Algebra" — 8 important questions with detailed answers for CBSE board exam…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Mathematics — Vector Algebra. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Vector Algebra — All 8 Questions Solved
Q1. Find the magnitude of the vector a = 3î + 4ĵ.
To find the magnitude of a = 3î + 4ĵ:
Step 1: Use the magnitude formula.
For a vector a = xî + yĵ + zk̂,
|a| = √(x² + y² + z²)
Step 2: Identify components.
Here: x = 3, y = 4, z = 0
Step 3: Calculate.
|a| = √(3² + 4² + 0²)
= √(9 + 16)
= √25
= 5
Final Answer: |a| = 5 units
Q2. Find the dot product of vectors a = 2î + 3ĵ + k̂ and b = î + 2ĵ - k̂.
To find a · b where a = 2î + 3ĵ + k̂ and b = î + 2ĵ - k̂:
Step 1: Use dot product formula.
a · b = (a₁î + a₂ĵ + a₃k̂) · (b₁î + b₂ĵ + b₃k̂)
= a₁b₁ + a₂b₂ + a₃b₃
Step 2: Identify components.
a₁ = 2, a₂ = 3, a₃ = 1
b₁ = 1, b₂ = 2, b₃ = -1
Step 3: Calculate.
a · b = (2)(1) + (3)(2) + (1)(-1)
= 2 + 6 - 1
= 7
Final Answer: a · b = 7
Q3. Find the cross product a × b where a = î + 2ĵ + 3k̂ and b = 2î + 3ĵ + k̂.
To find a × b where a = î + 2ĵ + 3k̂ and b = 2î + 3ĵ + k̂:
Step 1: Use the determinant formula.
a × b = |î ĵ k̂ |
|1 2 3 |
|2 3 1 |
Step 2: Expand the determinant.
a × b = î(2×1 - 3×3) - ĵ(1×1 - 3×2) + k̂(1×3 - 2×2)
= î(2 - 9) - ĵ(1 - 6) + k̂(3 - 4)
= î(-7) - ĵ(-5) + k̂(-1)
= -7î + 5ĵ - k̂
Final Answer: a × b = -7î + 5ĵ - k̂
Q4. Find the angle between vectors a = î + ĵ and b = ĵ + k̂.
To find the angle θ between a = î + ĵ and b = ĵ + k̂:
Step 1: Calculate the dot product a · b.
a · b = (1)(0) + (1)(1) + (0)(1) = 1
Step 2: Calculate magnitudes.
|a| = √(1² + 1² + 0²) = √2
|b| = √(0² + 1² + 1²) = √2
Step 3: Use the formula cos θ = (a · b)/(|a||b|).
cos θ = 1/(√2 × √2)
= 1/2
Step 4: Find θ.
θ = cos⁻¹(1/2)
= π/3 or 60°
Final Answer: θ = π/3 (or 60°)
Q5. Find the scalar projection of a = 2î + 3ĵ - k̂ on b = î + ĵ + k̂.
To find scalar projection of a on b:
Step 1: Use scalar projection formula.
Scalar projection of a on b = (a · b)/|b|
Step 2: Calculate a · b.
a · b = (2)(1) + (3)(1) + (-1)(1)
= 2 + 3 - 1
= 4
Step 3: Calculate |b|.
|b| = √(1² + 1² + 1²) = √3
Step 4: Calculate scalar projection.
Scalar projection = 4/√3
= 4√3/3
Final Answer: Scalar projection = 4/√3 = 4√3/3
Q6. Verify that vectors a = 2î + 3ĵ and b = -3î + 2ĵ are perpendicular.
To verify that a and b are perpendicular:
Step 1: Recall perpendicularity condition.
Two vectors are perpendicular if and only if a · b = 0
Step 2: Calculate the dot product.
a · b = (2)(-3) + (3)(2) + (0)(0)
= -6 + 6 + 0
= 0
Step 3: Conclusion.
Since a · b = 0, the vectors are perpendicular.
Final Answer: Yes, vectors a and b are perpendicular (verified)
Q7. Find the unit vector in the direction of a = 6î - 8ĵ.
To find the unit vector in the direction of a:
Step 1: Calculate the magnitude of a.
|a| = √(6² + (-8)²)
= √(36 + 64)
= √100
= 10
Step 2: Use the unit vector formula.
Unit vector = a/|a|
Step 3: Calculate.
unit vector = (6î - 8ĵ)/10
= (6/10)î - (8/10)ĵ
= (3/5)î - (4/5)ĵ
= 0.6î - 0.8ĵ
Final Answer: û = (3/5)î - (4/5)ĵ or 0.6î - 0.8ĵ
Q8. Find the area of the triangle with vertices A(1, 2, 3), B(2, 3, 4), and C(4, 5, 6).
To find the area of triangle ABC:
Step 1: Find vectors AB and AC.
AB = B - A = (2-1)î + (3-2)ĵ + (4-3)k̂ = î + ĵ + k̂
AC = C - A = (4-1)î + (5-2)ĵ + (6-3)k̂ = 3î + 3ĵ + 3k̂
Step 2: Calculate the cross product AB × AC.
AB × AC = |î ĵ k̂ |
|1 1 1 |
|3 3 3 |
= î(1×3 - 1×3) - ĵ(1×3 - 1×3) + k̂(1×3 - 1×3)
= î(3 - 3) - ĵ(3 - 3) + k̂(3 - 3)
= 0î + 0ĵ + 0k̂
= 0
Step 3: Find the magnitude.
|AB × AC| = |0| = 0
Step 4: Calculate area.
Area = (1/2)|AB × AC| = (1/2) × 0 = 0
Step 5: Interpretation.
Since the cross product is zero, the vectors AB and AC are parallel/collinear.
This means points A, B, C are collinear (lie on the same line).
Final Answer: Area = 0 (points are collinear)
Vector Algebra — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. If $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$, what is the magnitude of $\vec{a}$?
- A. 7
- B. $\sqrt{13}$
- C. 11
- D. $\sqrt{49}$
Answer: A. 7 — The magnitude of a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$. Here, $|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
MCQ 2. Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$.
- A. $\hat{i} + \hat{j} + \hat{k}$
- B. $\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$
- C. $\frac{1}{\sqrt{3}}(\hat{i} + 2\hat{j} + 2\hat{k})$
- D. $\frac{1}{9}(\hat{i} + 2\hat{j} + 2\hat{k})$
Answer: B. $\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$ — A unit vector in the direction of $\vec{a}$ is given by $\frac{\vec{a}}{|\vec{a}|}$. First, calculate the magnitude: $|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$. So the unit vector is $\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$.
MCQ 3. If $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, find the vector $2\vec{a} - \vec{b}$.
- A. $3\hat{i} - 4\hat{j} + 7\hat{k}$
- B. $3\hat{i} + 4\hat{j} + 7\hat{k}$
- C. $3\hat{i} - 4\hat{j} + 5\hat{k}$
- D. $3\hat{i} - 4\hat{j} - 7\hat{k}$
Answer: A. $3\hat{i} - 4\hat{j} + 7\hat{k}$ — First, calculate $2\vec{a} = 2(2\hat{i} - \hat{j} + 3\hat{k}) = 4\hat{i} - 2\hat{j} + 6\hat{k}$. Then, subtract $\vec{b}$: $2\vec{a} - \vec{b} = (4\hat{i} - 2\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = (4-1)\hat{i} + (-2-2)\hat{j} + (6-(-1))\hat{k} = 3\hat{i} - 4\hat{j} + 7\hat{k}$.
MCQ 4. The position vectors of points A and B are $\vec{a}$ and $\vec{b}$ respectively. Find the position vector of a point C which divides the line segment AB internally in the ratio 2:1.
- A. $\frac{2\vec{a} + \vec{b}}{3}$
- B. $\frac{\vec{a} + 2\vec{b}}{3}$
- C. $\frac{2\vec{a} - \vec{b}}{1}$
- D. $\frac{\vec{a} - 2\vec{b}}{-1}$
Answer: B. $\frac{\vec{a} + 2\vec{b}}{3}$ — The section formula for internal division states that if a point C divides the line segment joining points A (with position vector $\vec{a}$) and B (with position vector $\vec{b}$) in the ratio $m:n$, then the position vector of C is $\frac{n\vec{a} + m\vec{b}}{m+n}$. Here, $m=2$ and $n=1$, so the position vector is $\frac{1\cdot\vec{a} + 2\cdot\vec{b}}{2+1} = \frac{\vec{a} + 2\vec{b}}{3}$.
MCQ 5. Which of the following conditions implies that vectors $\vec{a}$ and $\vec{b}$ are collinear?
- A. $\vec{a} \cdot \vec{b} = 0$
- B. $\vec{a} + \vec{b} = \vec{0}$
- C. $\vec{a} = k\vec{b}$ for some scalar $k \ne 0$
- D. $|\vec{a}| = |\vec{b}|$
Answer: C. $\vec{a} = k\vec{b}$ for some scalar $k \ne 0$ — Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if there exists a non-zero scalar $k$ such that $\vec{a} = k\vec{b}$. This means they are parallel or anti-parallel.
MCQ 6. If $\vec{a} = \hat{i} + \lambda\hat{j} + 3\hat{k}$ and $\vec{b} = 4\hat{i} + 2\hat{j} - 4\hat{k}$ are perpendicular to each other, find the value of $\lambda$.
- A. 4
- B. -4
- C. 2
- D. -2
Answer: A. 4 — Two vectors are perpendicular if their dot product is zero. So, $\vec{a} \cdot \vec{b} = 0$. $(\hat{i} + \lambda\hat{j} + 3\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 4\hat{k}) = (1)(4) + (\lambda)(2) + (3)(-4) = 0$. This simplifies to $4 + 2\lambda - 12 = 0$, so $2\lambda - 8 = 0$, which gives $2\lambda = 8$, and $\lambda = 4$.
MCQ 7. Find the angle between the vectors $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$.
- A. $0^\circ$
- B. $30^\circ$
- C. $45^\circ$
- D. $60^\circ$
Answer: D. $60^\circ$ — The cosine of the angle $\theta$ between two vectors is given by $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. Here, $\vec{a} \cdot \vec{b} = (1)(0) + (1)(1) + (0)(1) = 1$. $|\vec{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}$. $|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$. So, $\cos\theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$. Therefore, $\theta = 60^\circ$.
MCQ 8. Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
- A. $\frac{10}{\sqrt{6}}$
- B. $\frac{10}{6}$
- C. $\frac{6}{10}$
- D. $\frac{\sqrt{6}}{10}$
Answer: A. $\frac{10}{\sqrt{6}}$ — The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. $\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$. $|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$. So, the projection is $\frac{10}{\sqrt{6}}$.
MCQ 9. Find the area of the parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} + \hat{k}$.
- A. $5\sqrt{3}$
- B. $5\sqrt{2}$
- C. $5\sqrt{6}$
- D. $10\sqrt{2}$
Answer: C. $5\sqrt{6}$ — The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}|$. First, calculate the cross product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 1 \end{vmatrix} = \hat{i}(1 - 6) - \hat{j}(3 - (-2)) + \hat{k}(-9 - 1) = -5\hat{i} - 5\hat{j} - 10\hat{k}$. Next, calculate the magnitude: $|\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + (-5)^2 + (-10)^2} = \sqrt{25 + 25 + 100} = \sqrt{150}$. Finally, simplify the radical: $\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}$ square units.
MCQ 10. If the vertices of a triangle are A(0,0,0), B(1,1,1), and C(0,2,3), find its area.
- A. $\frac{\sqrt{14}}{2}$
- B. $\sqrt{14}$
- C. $\frac{\sqrt{10}}{2}$
- D. $\frac{\sqrt{6}}{2}$
Answer: A. $\frac{\sqrt{14}}{2}$ — The area of a triangle with vertices A, B, C is given by $\frac{1}{2}|\vec{AB} \times \vec{AC}|$. First, find vectors $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = (1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k} = \hat{i} + \hat{j} + \hat{k}$. $\vec{AC} = (0-0)\hat{i} + (2-0)\hat{j} + (3-0)\hat{k} = 2\hat{j} + 3\hat{k}$. Next, calculate their cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-0) + \hat{k}(2-0) = \hat{i} - 3\hat{j} + 2\hat{k}$. Finally, find the magnitude and area: $|\vec{AB} \times \vec{AC}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1+9+4} = \sqrt{14}$. The area is $\frac{1}{2}\sqrt{14}$ square units.
MCQ 11. Find the volume of the parallelepiped whose co-terminus edges are represented by the vectors $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, and $\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}$.
- A. 7 cubic units
- B. 14 cubic units
- C. 21 cubic units
- D. 28 cubic units
Answer: A. 7 cubic units — The volume of a parallelepiped with co-terminus edges $\vec{a}$, $\vec{b}$, $\vec{c}$ is given by the absolute value of their scalar triple product, $|[\vec{a}, \vec{b}, \vec{c}]| = \left| \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} \right|$. This evaluates to $|2(4-1) - (-3)(2-(-3)) + 4(-1-6)| = |2(3) + 3(5) + 4(-7)| = |6 + 15 - 28| = |-7| = 7$ cubic units.
MCQ 12. For what value of $\lambda$ are the vectors $\vec{a} = \hat{i} + 3\hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} - \hat{k}$, and $\vec{c} = \lambda\hat{i} + 7\hat{j} + 3\hat{k}$ coplanar?
- A. 0
- B. 1
- C. 2
- D. -1
Answer: A. 0 — Three vectors are coplanar if their scalar triple product is zero. So, $\begin{vmatrix} 1 & 3 & 1 \\ 2 & -1 & -1 \\ \lambda & 7 & 3 \end{vmatrix} = 0$. Expanding the determinant: $1((-1)(3) - (-1)(7)) - 3(2 \times 3 - (-1)\lambda) + 1(2 \times 7 - (-1)\lambda) = 0$. This simplifies to $1(-3 + 7) - 3(6 + \lambda) + 1(14 + \lambda) = 0$, which means $4 - 18 - 3\lambda + 14 + \lambda = 0$, so $-2\lambda = 0 \Rightarrow \lambda = 0$.
MCQ 13. If a line makes angles $90^\circ, 60^\circ, 30^\circ$ with the positive x, y, z-axes respectively, what are its direction cosines?
- A. $(0, \frac{1}{2}, \frac{\sqrt{3}}{2})$
- B. $(\frac{1}{2}, 0, \frac{\sqrt{3}}{2})$
- C. $(0, \frac{\sqrt{3}}{2}, \frac{1}{2})$
- D. $(\frac{\sqrt{3}}{2}, \frac{1}{2}, 0)$
Answer: A. $(0, \frac{1}{2}, \frac{\sqrt{3}}{2})$ — The direction cosines are given by $\cos\alpha$, $\cos\beta$, $\cos\gamma$, where $\alpha, \beta, \gamma$ are the angles the line makes with the positive x, y, z-axes respectively. Here, $\alpha = 90^\circ$, $\beta = 60^\circ$, $\gamma = 30^\circ$. So, $\cos 90^\circ = 0$, $\cos 60^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$. The direction cosines are $(0, \frac{1}{2}, \frac{\sqrt{3}}{2})$.
MCQ 14. If $l, m, n$ are the direction cosines of a line, then which of the following is always true?
- A. $l+m+n=1$
- B. $l^2+m^2+n^2=0$
- C. $l^2+m^2+n^2=1$
- D. $l=m=n$
Answer: C. $l^2+m^2+n^2=1$ — The sum of the squares of the direction cosines of any line is always equal to 1, i.e., $l^2 + m^2 + n^2 = 1$. This is a fundamental property derived from the definition of direction cosines.
MCQ 15. Find the direction ratios of the line passing through the points A(1,-2,3) and B(3,4,-5).
- A. $(2, 6, -8)$
- B. $(4, 2, -2)$
- C. $(-2, -6, 8)$
- D. $(2, -6, 8)$
Answer: A. $(2, 6, -8)$ — The direction ratios of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1)$. Here, $(x_2-x_1) = 3-1 = 2$. $(y_2-y_1) = 4-(-2) = 6$. $(z_2-z_1) = -5-3 = -8$. So, the direction ratios are $(2, 6, -8)$.
MCQ 16. A vector has magnitude 7 and has direction ratios $(2, -3, 6)$. Find the vector.
- A. $2\hat{i} - 3\hat{j} + 6\hat{k}$
- B. $\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}$
- C. $\frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k})$
- D. $(2\hat{i} - 3\hat{j} + 6\hat{k})$ (This is redundant, but points to option A)
Answer: A. $2\hat{i} - 3\hat{j} + 6\hat{k}$ — A vector with direction ratios $(a,b,c)$ can be written as $k(a\hat{i} + b\hat{j} + c\hat{k})$ for some scalar $k$. The magnitude of $a\hat{i} + b\hat{j} + c\hat{k}$ is $\sqrt{a^2+b^2+c^2}$. Here, the direction ratios are $(2, -3, 6)$. The magnitude of the vector $2\hat{i} - 3\hat{j} + 6\hat{k}$ is $\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$. Since the desired magnitude is 7, the vector is simply $2\hat{i} - 3\hat{j} + 6\hat{k}$.
MCQ 17. A quantity that has both magnitude and direction is called a:
- A. Scalar
- B. Vector
- C. Tensor
- D. Constant
Answer: B. Vector — A vector is a quantity that possesses both magnitude and direction. Examples include displacement, velocity, and force.
MCQ 18. If A, B, C are the vertices of a triangle, then $\vec{AB} + \vec{BC} + \vec{CA}$ is equal to:
- A. $\vec{AC}$
- B. $\vec{CB}$
- C. $\vec{BA}$
- D. $\vec{0}$
Answer: D. $\vec{0}$ — According to the triangle law of vector addition, $\vec{AB} + \vec{BC} = \vec{AC}$. Therefore, $\vec{AB} + \vec{BC} + \vec{CA} = \vec{AC} + \vec{CA}$. Since $\vec{CA} = -\vec{AC}$, we have $\vec{AC} + (-\vec{AC}) = \vec{0}$.
MCQ 19. Which of the following is true for the scalar product (dot product) of two vectors $\vec{a}$ and $\vec{b}$?
- A. $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \sin\theta$
- B. $\vec{a} \cdot \vec{b}$ is a vector quantity
- C. $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
- D. $\vec{a} \cdot \vec{a} = 0$
Answer: C. $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ — The scalar product is commutative, meaning $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$. Option A is for cross product magnitude. Option B is incorrect as the dot product yields a scalar. Option D is incorrect as $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.
MCQ 20. Which of the following is NOT true for the vector product (cross product) of two vectors $\vec{a}$ and $\vec{b}$?
- A. $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$
- B. $\vec{a} \times \vec{b} = \vec{0}$ if $\vec{a}$ and $\vec{b}$ are parallel
- C. $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$
- D. $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{c} \times \vec{b}$
Answer: D. $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{c} \times \vec{b}$ — The vector product is distributive over addition but the order matters. The correct distributive property is $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \times \vec{c}$. Option D changes the order for the second term, so $\vec{c} \times \vec{b}$ should be $-(\vec{b} \times \vec{c})$.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.